Statistics and Probability Practice Questions — JAMB UTME — Page 7

Question 121Question →

The frequency table below shows the fuel efficiency, measured in $\text{km/L}$ , for a fleet of $50$ delivery vans operated by a logistics company:

Fuel Efficiency (km/L)	Frequency ( $f$ )
10 – 14	8
15 – 19	12
20 – 24	20
25 – 29	10

What is the mean fuel efficiency of the fleet, in $\text{km/L}$ ?

Show answer & explanation

Answer: 20.2

Answer

The mean fuel efficiency of the fleet is 20.2 km/L.

The estimated mean is found by calculating the midpoint of each class interval, multiplying each midpoint by its class frequency, summing these products to get 1010, and dividing by the total frequency of 50, resulting in 20.2 km/L.

Step-by-Step Solution

1

Determine the midpoint (x) for each class interval.

Midpoints are 12 for 10–14, 17 for 15–19, 22 for 20–24, and 27 for 25��29.

For grouped frequency data, each class interval is represented by its midpoint x = (lower limit + upper limit) / 2.

2

Compute the product of frequency and midpoint (f * x) for each interval.

Products: (8 * 12) = 96, (12 * 17) = 204, (20 * 22) = 440, (10 * 27) = 270.

To calculate the total contribution of values from each class interval.

3

Sum all products (sum of f * x) and divide by the total number of delivery vans (sum of f).

sum of f * x = 96 + 204 + 440 + 270 = 1010. sum of f = 50. Mean = 1010 / 50 = 20.2.

The formula for the estimated mean of grouped data is mean = (sum of f * x) / (sum of f).

Key Concept

Measures of Central Tendency for Grouped Data (Grouped Mean)

Question 122Question →

A laboratory technician recorded the temperature changes (in °C) of a chemical reaction across five trials as $3$ , $6$ , $7$ , $9$ , and $15$ . What is the variance of this set of data?

Show answer & explanation

Answer: 16

Answer

The variance of the temperature changes is 16.

The mean of the given numbers is

\bar{x} = \frac{3 + 6 + 7 + 9 + 15}{5} = 8

. The squared deviations from the mean are

(3-8)^2 = 25

,

(6-8)^2 = 4

,

(7-8)^2 = 1

,

(9-8)^2 = 1

, and

(15-8)^2 = 49

. Summing these squared deviations gives

25 + 4 + 1 + 1 + 49 = 80

. Dividing by the number of observations

N = 5

yields the variance:

\frac{80}{5} = 16

.

Step-by-Step Solution

1

Calculate the arithmetic mean (\bar{x}) of the dataset.

\bar{x} = \frac{3 + 6 + 7 + 9 + 15}{5} = 8

The mean is required to determine the deviation of each individual value from the central value.

2

Compute the square of the deviation for each data point from the mean.

(3-8)^2 = 25, (6-8)^2 = 4, (7-8)^2 = 1, (9-8)^2 = 1, (15-8)^2 = 49

Squaring deviations ensures all values are positive and emphasizes larger departures from the mean.

3

Sum the squared deviations and divide by the total number of observations (N = 5).

\text{Variance} = \frac{25 + 4 + 1 + 1 + 49}{5} = \frac{80}{5} = 16

Variance measures the average of the squared deviations from the mean.

Key Concept

Population Variance for Ungrouped Data

Question 123Question →

A fair six-sided die is rolled $150$ times in a probability experiment. If the number $4$ lands face up $35$ times, what is the positive difference between the experimental probability and the theoretical probability of rolling a $4$ ?

\frac{1}{15}

\frac{7}{30}

\frac{1}{6}

\frac{2}{5}

Show answer & explanation

Answer: $\frac{1}{15}$

Answer

The positive difference between the experimental probability and the theoretical probability is

\frac{1}{15}

.

The theoretical probability of getting a 4 on a fair die is

\frac{1}{6}

, which simplifies to

\frac{5}{30}

. The experimental probability from

35

successes out of

150

trials is

\frac{35}{150} = \frac{7}{30}

. Taking the positive difference yields

\frac{7}{30} - \frac{5}{30} = \frac{2}{30} = \frac{1}{15}

.

Step-by-Step Solution

1

Calculate the theoretical probability of rolling a 4 on a fair six-sided die

P(\text{theoretical}) = \frac{1}{6}

A fair six-sided die has 6 equally likely outcomes, and exactly one outcome is 4.

2

Calculate the experimental probability based on the experiment results

P(\text{experimental}) = \frac{35}{150} = \frac{7}{30}

Experimental probability is the ratio of successful trials (35) to the total number of trials (150).

3

Subtract the theoretical probability from the experimental probability to find the positive difference

\text{Difference} = \frac{7}{30} - \frac{1}{6} = \frac{7}{30} - \frac{5}{30} = \frac{2}{30} = \frac{1}{15}

Converting to a common denominator of 30 allows straightforward subtraction.

Key Concept

Comparing experimental probability (relative frequency from trials) to theoretical probability (expected frequency based on equally likely outcomes).

Question 124Question →

A convex polygon has $44$ diagonals. What is the total number of distinct triangles that can be formed by joining any three vertices of this polygon?

Show answer & explanation

Answer: 165

Answer

165

The number of diagonals of an

n

-sided convex polygon is given by

\binom{n}{2} - n = \frac{n(n-3)}{2}

. Setting this equal to

44

yields

n(n-3) = 88

, which simplifies to

n^2 - 3n - 88 = 0

. Factoring gives

(n-11)(n+8) = 0

, so

n = 11

. The total number of distinct triangles formed by choosing any 3 vertices from an 11-sided polygon is

\binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165

.

Step-by-Step Solution

1

Set up the equation for the number of diagonals in terms of the number of vertices

n

\frac{n(n-3)}{2} = 44

Choosing any 2 vertices from

n

vertices gives

\binom{n}{2}

total connecting line segments. Subtracting the

n

boundary sides leaves the diagonals.

2

Solve the quadratic equation for

n

n^2 - 3n - 88 = 0 \implies (n-11)(n+8) = 0 \implies n = 11

A polygon must have a positive integer number of vertices, so

n = 11

.

3

Compute the number of distinct triangles using combinations

\binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165

In a convex polygon, no three vertices are collinear, so every unique combination of 3 vertices forms a distinct triangle.

Key Concept

Application of combinations to geometry (polygon diagonals and triangle selection)

Estimated Time:2m 0s

Question 125Question →

In an agricultural trial, two independent seed varieties, $V_1$ and $V_2$ , have germination probabilities $P(V_1) = p$ and $P(V_2) = p + 0.20$ . If the probability that at least one seed variety germinates is $0.76$ , what is the probability that only variety $V_2$ germinates?

0.36

0.48

0.52

0.24

Show answer & explanation

Answer: 0.36

Answer

The probability that only variety

V_2

germinates is

0.36

.

Using the addition law

P(V_1 \cup V_2) = P(V_1) + P(V_2) - P(V_1)P(V_2)

gives

0.76 = 2p + 0.20 - (p^2 + 0.20p)

. Solving

p^2 - 1.80p + 0.56 = 0

gives

p = 0.40

. Thus

P(V_1) = 0.40

and

P(V_2) = 0.60

. The probability that only variety

V_2

germinates is

P(V_2) \cdot P(V_1') = 0.60 \times 0.60 = 0.36

.

Step-by-Step Solution

1

Set up the probability addition law for independent events.

P(V_1 \cup V_2) = P(V_1) + P(V_2) - P(V_1 \cap V_2)

. Since

V_1

and

V_2

are independent,

P(V_1 \cap V_2) = P(V_1) \cdot P(V_2) = p(p + 0.20)

.

Independent events allow the intersection probability to be expressed as the product of their individual probabilities.

2

Substitute the given values into the addition law and solve for

p

.

0.76 = p + (p + 0.20) - p(p + 0.20) \implies p^2 - 1.80p + 0.56 = 0 \implies (p - 0.40)(p - 1.40) = 0

. Since

p \le 1

,

p = 0.40

.

Formulating a quadratic equation yields the value of

p

within valid probability bounds.

3

Calculate individual probabilities

P(V_1)

and

P(V_2)

.

P(V_1) = 0.40

and

P(V_2) = 0.40 + 0.20 = 0.60

.

Knowing

p

gives the exact germination probabilities for both varieties.

4

Find the probability that only variety

V_2

germinates.

P(\text{only } V_2) = P(V_2 \cap V_1') = P(V_2) \times [1 - P(V_1)] = 0.60 \times (1 - 0.40) = 0.60 \times 0.60 = 0.36

.

Only

V_2

germinating means

V_2

germinates and

V_1

fails to germinate.

Key Concept

Probability laws for independent compound events

Question 126Question →

The frequency distribution table below shows the daily profits (in thousands of Naira, $\text{₦}$ ) recorded by a sample of $40$ small-scale market traders:

Daily Profit ( $\text{₦'000}$ )	Frequency ( $f$ )
$10 - 14$	$6$
$15 - 19$	$10$
$20 - 24$	$12$
$25 - 29$	$8$
$30 - 34$	$4$

What is the mean daily profit of the traders, in thousands of Naira?

Show answer & explanation

Answer: 21.25

Answer

The mean daily profit of the traders is

21.25

thousand Naira.

The mean daily profit is found by dividing the sum of the product of each class midpoint and its frequency (

\sum fx = 850

) by the total frequency (

\sum f = 40

), yielding

21.25

.

Step-by-Step Solution

1

Find the class midpoints (

x

) for each interval

Midpoints are

12

,

17

,

22

,

27

, and

32

.

For grouped data, each class interval is represented by its midpoint.

2

Calculate the product of frequency and midpoint (

f \cdot x

) for each interval

Products are

72

,

170

,

264

,

216

, and

128

.

Multiplying class midpoint by class frequency estimates the sum of values within that class.

3

Calculate total frequency (

\sum f

) and total sum of products (

\sum fx

)

\sum f = 40

and

\sum fx = 850

.

The sum of frequencies gives the total number of observations, and the sum of products gives the estimated grand total.

4

Apply the grouped mean formula

\bar{x} = \frac{\sum fx}{\sum f}

\bar{x} = \frac{850}{40} = 21.25.

Dividing total sum by total frequency gives the mean value.

Key Concept

Measures of Central Tendency for Grouped Data - Mean

Question 127Question →

A student library committee is selecting $5$ distinct books from a shelf containing $7$ novel titles and $5$ biography titles. If $2$ specific novel titles are mutually exclusive (they cannot both be selected together in the same combination), in how many ways can the selection of $5$ books be made such that at least $3$ novel titles are included?

426

436

516

546

Show answer & explanation

Answer: 436

Answer

436

To find the number of valid ways, we first find the total number of ways to select at least

3

novels from

7

novels and

5

biographies, which equals

546

. Next, we determine how many of these combinations contain both of the restricted novels while still having at least

3

novels overall. There are

110

such invalid selections. Subtracting

110

from

546

gives

436

valid ways.

Step-by-Step Solution

1

Calculate the total combinations with at least 3 novels without any restriction.

Case 1 (

3

Novels,

2

Biographies):

\binom{7}{3} \times \binom{5}{2} = 35 \times 10 = 350

.
Case 2 (

4

Novels,

1

Biography):

\binom{7}{4} \times \binom{5}{1} = 35 \times 5 = 175

.
Case 3 (

5

Novels,

0

Biographies):

\binom{7}{5} \times \binom{5}{0} = 21 \times 1 = 21

.
Total unrestricted combinations =

350 + 175 + 21 = 546

.

Establishing the total pool of choices that satisfy the constraint of selecting at least 3 novels.

2

Calculate the invalid combinations where both restricted novels are selected together AND at least 3 novels are included.

If both restricted novels are selected, we have already chosen

2

novels. To reach a total of

5

books with at least

3

novels, we must pick

3

additional books from the remaining

10

books (

5

remaining novels and

5

biographies), EXCLUDING the scenario where all

3

additional books are biographies (which would leave us with only

2

novels in total).
Combinations of

3

books from

10

remaining books =

\binom{10}{3} = 120

.
Combinations of

3

biographies from

5

biographies =

\binom{5}{3} = 10

.
Invalid combinations =

120 - 10 = 110

.

To apply the mutual exclusion constraint correctly, we must subtract only those invalid selections that also satisfy the condition of having at least 3 novels.

3

Subtract invalid combinations from total unrestricted combinations.

Valid combinations =

546 - 110 = 436

.

Subtracting the forbidden overlapping outcomes yields the net valid choices.

Key Concept

Combinations with multiple constraints and mutual exclusion

Question 128Question →

A music festival coordinator needs to select $4$ bands to perform from a pool of $8$ available bands. If $2$ specific bands insist on either both being selected or neither being selected, in how many different ways can the $4$ bands be chosen?

30

60

70

390

Show answer & explanation

Answer: $30$

Answer

The total number of ways to choose the bands under the given condition is

30

.

To satisfy the condition that the two specific bands are either both chosen or neither chosen, we evaluate two distinct cases. Case 1 (both chosen) requires choosing

2

additional bands from the remaining

6

, yielding

{}^6C_2 = 15

ways. Case 2 (neither chosen) requires choosing all

4

bands from the remaining

6

, yielding

{}^6C_4 = 15

ways. Adding both cases gives

15 + 15 = 30

ways.

Step-by-Step Solution

1

Analyze Case 1: Both specific bands are selected.

If both specific bands are included, we only need to select

2

more bands from the remaining

6

bands. The number of ways is

{}^6C_2 = \frac{6 \times 5}{2 \times 1} = 15

.

Since order does not matter in forming a group of performers, we use combinations.

2

Analyze Case 2: Neither of the specific bands is selected.

If neither of the

2

specific bands is chosen, all

4

bands must be selected from the remaining

6

bands. The number of ways is

{}^6C_4 = {}^6C_2 = 15

.

Excluding the

2

specific bands leaves

6

candidate bands to choose

4

from.

3

Sum the mutually exclusive cases.

\text{Total ways} = 15 + 15 = 30.

The two scenarios are disjoint, so the addition principle applies.

Key Concept

Combinations with conditional restrictions

Estimated Time:1m 30s

Question 129Question →

A bag contains $x$ red balls, $12$ blue balls, and $8$ green balls. The theoretical probability of selecting a green ball at random from the bag is $\frac{1}{4}$ . In an experiment where a ball is drawn with replacement $240$ times, a blue ball is recorded $102$ times. What is the positive difference between the experimental relative frequency and the theoretical probability of drawing a blue ball?

\frac{1}{20}

\frac{7}{40}

\frac{3}{8}

\frac{17}{40}

Show answer & explanation

Answer: $\frac{1}{20}$

Answer

The positive difference between the experimental relative frequency and the theoretical probability of drawing a blue ball is

\frac{1}{20}

.

The total number of balls in the container is determined from the theoretical probability of green balls:

\frac{8}{N} = \frac{1}{4} \implies N = 32

. The theoretical probability of drawing a blue ball is

\frac{12}{32} = \frac{3}{8} = \frac{15}{40}

. The experimental relative frequency is

\frac{102}{240} = \frac{17}{40}

. Taking the positive difference yields

\frac{17}{40} - \frac{15}{40} = \frac{2}{40} = \frac{1}{20}

.

Step-by-Step Solution

1

Calculate total number of balls in the bag

Total balls

N = 32

Given theoretical

P(\text{Green}) = \frac{8}{N} = \frac{1}{4}

, solving for

N

yields

N = 32

.

2

Determine theoretical probability of selecting a blue ball

P(\text{Blue}) = \frac{3}{8} = \frac{15}{40}

There are

12

blue balls out of

32

total balls, so

P(\text{Blue}) = \frac{12}{32} = \frac{3}{8}

.

3

Calculate experimental relative frequency of selecting a blue ball

Relative frequency =

\frac{17}{40}

Blue was drawn

102

times in

240

trials, so

\frac{102}{240} = \frac{17}{40}

.

4

Find the positive difference between experimental relative frequency and theoretical probability

Difference =

\frac{1}{20}

Subtracting

\frac{15}{40}

from

\frac{17}{40}

gives

\frac{2}{40} = \frac{1}{20}

.

Key Concept

Experimental relative frequency vs theoretical probability comparison

Estimated Time:2m 0s

Question 130Question →

Two candidates, $X$ and $Y$ , sit for an entrance examination independently. If the probability that candidate $X$ passes is $\frac{4}{7}$ and the probability that candidate $Y$ passes is $\frac{1}{3}$ , what is the probability that at least one of them passes the examination?

\frac{19}{21}

\frac{5}{7}

\frac{4}{21}

\frac{8}{21}

Show answer & explanation

Answer: $\frac{5}{7}$

Answer

The probability that at least one candidate passes the examination is

\frac{5}{7}

.

The probability that at least one candidate passes is given by the union of independent events

P(X \cup Y) = P(X) + P(Y) - P(X)P(Y)

. Substituting the given values yields

\frac{4}{7} + \frac{1}{3} - \left(\frac{4}{7} \cdot \frac{1}{3}\right) = \frac{15}{21} = \frac{5}{7}

. Alternatively, using the complement rule:

1 - P(X')P(Y') = 1 - \left(1 - \frac{4}{7}\right)\left(1 - \frac{1}{3}\right) = 1 - \left(\frac{3}{7} \cdot \frac{2}{3}\right) = 1 - \frac{2}{7} = \frac{5}{7}

.

Step-by-Step Solution

1

Identify given probabilities and state independence condition

P(X) = \frac{4}{7}

and

P(Y) = \frac{1}{3}

. Since

X

and

Y

are independent events,

P(X \cap Y) = P(X) \times P(Y)

.

Independent events allow the joint probability of both events occurring to be calculated as the product of their individual probabilities.

2

Calculate the intersection probability

P(X \cap Y)

P(X \cap Y) = \frac{4}{7} \times \frac{1}{3} = \frac{4}{21}

.

The intersection gives the probability that both candidate

X

and candidate

Y

pass.

3

Apply the general addition law of probability

P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)

P(X \cup Y) = \frac{4}{7} + \frac{1}{3} - \frac{4}{21} = \frac{12}{21} + \frac{7}{21} - \frac{4}{21} = \frac{15}{21} = \frac{5}{7}

.

The union of two events represents the event that at least one of them occurs.

Key Concept

Probability Laws for Compound and Independent Events

Alternative Method

Using the complementary law of probability:

P(\text{at least one passes}) = 1 - P(\text{neither passes}) = 1 - P(X')P(Y') = 1 - \left(1 - \frac{4}{7}\right)\left(1 - \frac{1}{3}\right) = 1 - \left(\frac{3}{7} \times \frac{2}{3}\right) = 1 - \frac{2}{7} = \frac{5}{7}

.

Estimated Time:1m 30s

Question 131Question →

The mean of a set of $7$ numbers arranged in ascending order is $16$ . If the median of the set is $16$ and the mean of the smallest $3$ numbers is $11$ , what is the mean of the largest $3$ numbers?

Show answer & explanation

Answer: 21

Answer

The mean of the largest 3 numbers is 21.

For a set of 7 ordered numbers, the median is the 4th number, which is given as 16. The total sum of all 7 numbers is 7 × 16 = 112. The sum of the 3 smallest numbers is 3 × 11 = 33. Subtracting the 3 smallest numbers and the median from the total sum leaves the sum of the 3 largest numbers: 112 - 33 - 16 = 63. Dividing 63 by 3 gives a mean of 21.

Step-by-Step Solution

1

Find the total sum of the set of 7 numbers

Sum = 7 × 16 = 112

The mean of a set is the total sum divided by the number of elements.

2

Determine the median value and the sum of the 3 smallest numbers

Median (4th number) = 16, Sum of 3 smallest = 3 × 11 = 33

For an ordered set of 7 numbers, the 4th number is the median, and the mean of the first 3 numbers gives their sum when multiplied by 3.

3

Calculate the sum of the 3 largest numbers

Sum of 3 largest = 112 - 33 - 16 = 63

The total sum is the sum of the 3 smallest numbers, the median (4th number), and the 3 largest numbers.

4

Compute the mean of the 3 largest numbers

Mean = 63 / 3 = 21

Dividing the sum of the 3 largest numbers by 3 gives their arithmetic mean.

Key Concept

Arithmetic Mean and Median of Ungrouped Data Subgroups

Question 132Question →

A committee of $5$ members is to be selected from $6$ doctors and $4$ nurses. In how many ways can this committee be formed if it must include at least $3$ doctors?

186

120

720

3600

Show answer & explanation

Answer: 186

Answer

186 ways

The required committee must contain at least 3 doctors out of 5 total members. The three possible scenarios are: 3 doctors and 2 nurses (

^6C_3 \times ^4C_2 = 120

), 4 doctors and 1 nurse (

^6C_4 \times ^4C_1 = 60

), and 5 doctors and 0 nurses (

^6C_5 \times ^4C_0 = 6

). Summing these gives

120 + 60 + 6 = 186

.

Step-by-Step Solution

1

Identify the possible valid committee compositions given the condition 'at least 3 doctors'.

Three mutually exclusive cases exist for a 5-member committee: (3 doctors, 2 nurses), (4 doctors, 1 nurse), or (5 doctors, 0 nurses).

The total size of the committee is 5, so selecting more doctors reduces the required number of nurses.

2

Calculate the combinations for Case 1: 3 doctors and 2 nurses.

^6C_3 \times ^4C_2 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{4 \times 3}{2 \times 1} = 20 \times 6 = 120

Order of selection within each group does not matter.

3

Calculate the combinations for Case 2: 4 doctors and 1 nurse.

^6C_4 \times ^4C_1 = \frac{6 \times 5}{2 \times 1} \times 4 = 15 \times 4 = 60

Selecting 4 doctors out of 6 is equivalent to choosing which 2 doctors to exclude.

4

Calculate the combinations for Case 3: 5 doctors and 0 nurses.

^6C_5 \times ^4C_0 = 6 \times 1 = 6

Choosing 5 doctors out of 6 yields 6 possibilities, and choosing 0 nurses yields 1.

5

Sum the number of ways from all valid cases.

120 + 60 + 6 = 186

By the addition principle of counting, the total ways to form the committee is the sum of ways across mutually exclusive cases.

Key Concept

Combinations with constraints (addition and multiplication principles)

Question 133Question →

A pie chart illustrates the distribution of undergraduate students enrolled across four faculties at a university: Arts, Science, Law, and Medicine. The central angles for the sectors representing Arts, Science, and Law are $120^\circ$ , $90^\circ$ , and $75^\circ$ , respectively. If $300$ students are enrolled in the Faculty of Medicine, what is the total number of students enrolled in the university?

Show answer & explanation

Answer: 1440

Answer

The total number of students enrolled in the university is 1440.

The total sum of central angles in any pie chart is

360^\circ

. Subtracting the given angles for Arts (

120^\circ

), Science (

90^\circ

), and Law (

75^\circ

) from

360^\circ

gives the sector angle for Medicine as

75^\circ

. Since

75^\circ

represents

300

students, each degree represents

\frac{300}{75} = 4

students. Multiplying

4

students per degree by the total

360^\circ

yields

1440

total students in the university.

Step-by-Step Solution

1

Determine the sector angle for Medicine.

Sector angle for Medicine =

75^\circ

The sum of central angles in a pie chart is

360^\circ

. Subtracting

120^\circ + 90^\circ + 75^\circ = 285^\circ

from

360^\circ

gives

75^\circ

.

2

Formulate the proportion relating sector angle to frequency.

\frac{75^\circ}{360^\circ} \times N = 300

, where

N

represents total students.

The fractional portion of the angle (

75^\circ

out of

360^\circ

) equals the fractional portion of the total student count (

300

out of

N

).

3

Calculate the total student population

N

.

N = \frac{300 \times 360}{75} = 1440

Dividing

300

by

75

yields

4

students per degree. Multiplying

4

by

360

gives

1440

students.

Key Concept

Pie Chart Sector Angle and Total Population Calculation

Question 134Question →

A wheel is divided into $10$ equal sectors numbered $1$ through $10$ . In an initial experiment, the wheel is spun $400$ times, and a prime number is recorded $180$ times. Additional spins are to be conducted, all of which result in non-prime numbers. How many additional spins must be performed so that the overall experimental relative frequency of landing on a prime number equals the theoretical probability of landing on a prime number?

50

100

200

450

Show answer & explanation

Answer: $50$

Answer

50 additional spins

The correct value is 50. The theoretical probability of landing on a prime number from 1 to 10 is 4/10 = 2/5, since the prime numbers are 2, 3, 5, and 7. With 180 prime outcomes in 400 spins, adding N non-prime spins yields an experimental relative frequency of 180 / (400 + N). Setting 180 / (400 + N) = 2/5 yields 800 + 2N = 900, giving N = 50.

Step-by-Step Solution

1

Determine the theoretical probability of landing on a prime number.

The prime numbers between

1

and

10

are

2, 3, 5,

and

7

. Thus, there are

4

favorable outcomes out of

10

, giving

P(\text{theoretical}) = \frac{4}{10} = \frac{2}{5}

.

Theoretical probability is the ratio of favorable outcomes to the total number of equally likely outcomes.

2

Formulate the expression for experimental relative frequency after

N

additional non-prime spins.

The number of prime occurrences remains

180

, while the total number of spins becomes

400 + N

. Thus,

P(\text{experimental}) = \frac{180}{400 + N}

.

Since all

N

additional spins produce non-prime numbers, the count of prime outcomes does not increase, but the total number of trials increases by

N

.

3

Equate the experimental relative frequency to the theoretical probability and solve for

N

.

\frac{180}{400 + N} = \frac{2}{5} \implies 2(400 + N) = 180 \times 5 \implies 800 + 2N = 900 \implies 2N = 100 \implies N = 50.

Setting the experimental relative frequency equal to the theoretical probability allows determining the exact number of additional non-prime spins needed.

Key Concept

Theoretical vs Experimental Probability

Estimated Time:2m 0s

Question 135Question →

In a mathematics examination paper consisting of $8$ questions, a candidate is required to answer $5$ questions in total. If the candidate must answer the first $2$ questions, in how many ways can the candidate select the remaining questions?

Show answer & explanation

Answer: 20

Answer

The candidate can select the questions in 20 ways.

Because the first 2 questions are mandatory, the choice is reduced to picking 3 additional questions from the remaining 6 questions. Since selection order is irrelevant, the number of distinct ways is $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

Step-by-Step Solution

1

Determine the remaining number of questions needed

3 questions

Since 2 out of the required 5 questions are compulsory, the candidate must choose 3 more.

2

Determine the available pool of remaining questions

6 questions

Subtracting the 2 compulsory questions from the 8 total questions leaves 6 questions available.

3

Calculate the combinations using the nCr formula

20 ways

The order in which questions are selected does not matter, so we use combinations: $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

Key Concept

Combinations with restricted or fixed choices

Question 136Question →

Find the value of $r$ such that ${^{8}P_r} = 6720$ .

Show answer & explanation

Answer: 5

Answer

The value of

r

is 5.

Expanding

{^{8}P_r}

into consecutive decreasing factors starting from 8 yields

8 \times 7 \times 6 \times 5 \times 4 = 6720

. Counting the number of factors multiplied (8, 7, 6, 5, 4) gives exactly 5 factors, so

r = 5

.

Step-by-Step Solution

1

Write the formula for permutation

{^{8}P_r}

as a product of descending integers.

{^{8}P_r} = 8 \times 7 \times 6 \times \dots \times (8 - r + 1)

By definition,

{^{n}P_r}

represents the product of

r

consecutive factors starting from

n

and decreasing by 1.

2

Perform sequential multiplication starting from 8 until reaching 6720.

Product of 5 factors:

8 \times 7 \times 6 \times 5 \times 4 = 6720

Multiplying factors gives:

8 \times 7 = 56

;

56 \times 6 = 336

;

336 \times 5 = 1680

;

1680 \times 4 = 6720

.

3

Count the number of terms in the product to find

r

.

r = 5

Since 5 consecutive integers were multiplied together to obtain 6720, the subset size

r

is 5.

Key Concept

Permutations of

n

distinct items taken

r

at a time

Estimated Time:1m 0s

Question 137Question →

Match each statistical chart term or parameter in Column A with its corresponding definition or formula in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Class boundary matches the exact continuous limit of a class interval; Frequency density matches the quotient of class frequency and class width; Cumulative frequency matches the running total of frequencies plotted on an ogive; Pie chart sector angle matches the central angle formula involving multiplication by 360 degrees.

Each chart parameter in Column A directly corresponds to its standard mathematical definition, formula, or geometric representation in Column B.

Step-by-Step Solution

1

Identify the statistical definition for continuous class intervals in histograms.

Class boundaries remove gaps between non-overlapping class limits.

Histograms require continuous real class boundaries on the horizontal axis.

2

Recall the formula for histogram bar height when class intervals vary.

Frequency density =

\frac{\text{Frequency}}{\text{Class Width}}

.

Bar area must remain proportional to frequency.

3

Determine the parameter used to plot an ogive curve.

Cumulative frequency tracks accumulated totals across upper class boundaries.

An ogive represents cumulative distribution.

4

Identify the angular calculation for circular charts.

Sector angle =

\frac{\text{Class Frequency}}{\text{Total Frequency}} \times 360^\circ

.

A complete pie chart represents 360 degrees.

Key Concept

Data Representation and Chart Properties

Question 138Question →

A student obtained a mean score of $62$ in $5$ subjects. After the score of a $6\text{th}$ subject was added, the overall mean score became $65$ . What is the score obtained in the $6\text{th}$ subject?

Show answer & explanation

Answer: 80

Answer

The score obtained in the 6th subject is 80.

The sum of scores for the first 5 subjects is

5 \times 62 = 310

. With the 6th subject included, the total sum of scores becomes

6 \times 65 = 390

. The score of the 6th subject is the difference between these two totals:

390 - 310 = 80

.

Step-by-Step Solution

1

Calculate the total score of the initial 5 subjects

310

The sum of scores for ungrouped data is equal to the number of data items multiplied by the mean: 5 × 62 = 310.

2

Calculate the total score of all 6 subjects after including the new score

390

The new total score is obtained by multiplying the new count of subjects by the new mean: 6 × 65 = 390.

3

Determine the score of the 6th subject

80

The difference between the total score of 6 subjects and the total score of 5 subjects gives the score of the 6th subject: 390 - 310 = 80.

Key Concept

Calculating a missing data value given the mean of ungrouped data before and after addition

Question 139Question →

The cumulative frequency distribution below shows the monthly electricity consumption (in kWh) recorded for $50$ households in a residential estate:

Electricity Usage (kWh)	Cumulative Frequency
$\leq 50$	5
$\leq 100$	15
$\leq 150$	35
$\leq 200$	45
$\leq 250$	50

Using linear interpolation, calculate the 60th percentile ( $P_{60}$ ) of electricity consumption in kWh.

Show answer & explanation

Answer: 137.5

Answer

137.5 kWh

The 60th percentile corresponds to the 30th observation out of 50 (

0.60 \times 50 = 30

). From the cumulative frequency table, the 30th value falls in the class interval between

100

and

150

. Using the lower boundary of

100

, previous cumulative frequency of

15

, interval frequency of

20

, and interval width of

50

, linear interpolation gives

100 + \frac{30 - 15}{20} \times 50 = 137.5

kWh.

Step-by-Step Solution

1

Determine the rank for the 60th percentile

Rank position = 30th household

The 60th percentile rank corresponds to

60\%

of the total sample size (

N = 50

), which is

0.60 \times 50 = 30

.

2

Identify the parameters of the percentile class interval

Lower boundary

L = 100

,

CF_{\text{prev}} = 15

, class frequency

f = 20

, width

w = 50

The cumulative frequency rises from 15 to 35 in the class interval

(100, 150]

, so the 30th value lies within this interval.

3

Calculate the value using linear interpolation

137.5 kWh

Substitute values into

P_{60} = 100 + \left(\frac{30 - 15}{20}\right) \times 50 = 100 + 37.5 = 137.5

kWh.

Key Concept

Linear interpolation for percentiles from a cumulative frequency distribution

Question 140Question →

The cumulative frequency distribution table below shows the mass of cocoa beans (in kg) harvested by $40$ smallholder farmers in a agricultural cooperative:

Mass (kg)	Cumulative Frequency
$\leq 20$	$4$
$\leq 30$	$12$
$\leq 40$	$28$
$\leq 50$	$36$
$\leq 60$	$40$

Using linear interpolation from the cumulative frequency table, what is the median mass (in kg) of cocoa beans harvested by the farmers?

Show answer & explanation

Answer: 35

Answer

The median mass of cocoa beans harvested by the farmers is

35.0\text{ kg}

.

The total number of farmers is

N = 40

. The median corresponds to the

\frac{40}{2} = 20^{\text{th}}

position. From the cumulative frequency table, the

20^{\text{th}}

item lies within the

30 - 40\text{ kg}

class interval. Using the grouped median formula

\text{Median} = L + \left(\frac{\frac{N}{2} - c.f.}{f}\right) \times w

, where

L = 30

,

c.f. = 12

,

f = 28 - 12 = 16

, and

w = 10

, we obtain

\text{Median} = 30 + \left(\frac{20 - 12}{16}\right) \times 10 = 35.0\text{ kg}

.

Step-by-Step Solution

1

Find the median position in the cumulative frequency distribution

Median position

= \frac{N}{2} = \frac{40}{2} = 20^{\text{th}}

item

The median corresponds to the 50th percentile, which is half of the total cumulative frequency

N = 40

.

2

Locate the median class interval and extract its statistical parameters

Median class interval is

30 - 40\text{ kg}

, with lower limit

L = 30\text{ kg}

, preceding cumulative frequency

c.f. = 12

, class frequency

f = 28 - 12 = 16

, and width

w = 10\text{ kg}

The cumulative frequency just below

20

is

12

(at upper boundary

30

), and at upper boundary

40

it rises to

28

.

3

Substitute values into the grouped data median formula

\text{Median} = 30 + \left(\frac{20 - 12}{16}\right) \times 10 = 30 + 5 = 35.0\text{ kg}

Linear interpolation estimates the exact position of the median within the median class interval.

Key Concept

Calculation of Median from Cumulative Frequency Data

Statistics and Probability

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