Statistics and Probability Alıştırma Soruları — JAMB UTME — Sayfa 3

Soru 41Soru →

Two candidates, Ada and Babatunde, independently sit for an aptitude test. The probability that Ada passes the test is $\frac{1}{3}$ and the probability that Babatunde passes is $\frac{2}{5}$ . What is the probability that at least one of them passes the test?

\frac{2}{15}

\frac{2}{5}

\frac{3}{5}

\frac{11}{15}

Cevabı ve açıklamayı göster

Cevap: $\frac{3}{5}$

Cevap

The probability that at least one candidate passes the test is

\frac{3}{5}

.

For independent events, the probability of both occurring is given by

P(A \cap B) = P(A) \cdot P(B) = \frac{1}{3} \cdot \frac{2}{5} = \frac{2}{15}

. Applying the addition law

P(A \cup B) = P(A) + P(B) - P(A \cap B)

yields

\frac{1}{3} + \frac{2}{5} - \frac{2}{15} = \frac{9}{15} = \frac{3}{5}

. Alternatively, calculating

1 - P(\text{neither passes}) = 1 - (1 - \frac{1}{3})(1 - \frac{2}{5}) = 1 - (\frac{2}{3} \cdot \frac{3}{5}) = 1 - \frac{2}{5} = \frac{3}{5}

gives the identical correct outcome.

Adım Adım Çözüm

1

Calculate the probability that both Ada and Babatunde pass the test.

P(\text{Ada} \cap \text{Babatunde}) = \frac{1}{3} \times \frac{2}{5} = \frac{2}{15}

Since the two events are independent, the multiplication law of probability applies:

P(A \cap B) = P(A) \times P(B)

.

2

Apply the general addition law of probability to find the union of the two events.

P(\text{Ada} \cup \text{Babatunde}) = P(\text{Ada}) + P(\text{Babatunde}) - P(\text{Ada} \cap \text{Babatunde})

The probability of at least one event occurring corresponds to the union of the compound events.

3

Substitute the probabilities and solve for the outcome.

P(\text{Ada} \cup \text{Babatunde}) = \frac{1}{3} + \frac{2}{5} - \frac{2}{15} = \frac{5 + 6 - 2}{15} = \frac{9}{15} = \frac{3}{5}

Find a common denominator (15) and simplify the resulting fraction to its lowest terms.

Anahtar Kavram

Probability Laws for Compound and Independent Events

Tahmini Süre:1m 30s

Soru 42Soru →

In a telecommunications network, two independent relay switches, $R_1$ and $R_2$ , operate during a data transmission. The probability that switch $R_1$ functions successfully is $0.80$ , and the probability that switch $R_2$ functions successfully is $0.75$ . What is the probability that at least one of the two switches functions successfully during the transmission?

Cevabı ve açıklamayı göster

Cevap: 0.95

Cevap

The probability that at least one switch functions successfully is 0.95.

Since the two switches operate independently, the probability of both failing is the product of their individual failure probabilities:

(1 - 0.80) \times (1 - 0.75) = 0.20 \times 0.25 = 0.05

. Therefore, the probability that at least one switch functions successfully is

1 - 0.05 = 0.95

. Alternatively, using the addition law for independent events:

P(R_1 \cup R_2) = P(R_1) + P(R_2) - P(R_1 \cap R_2) = 0.80 + 0.75 - (0.80 \times 0.75) = 1.55 - 0.60 = 0.95

.

Adım Adım Çözüm

1

Determine the probabilities of individual switch failure.

P(R_1') = 0.20 and P(R_2') = 0.25

The probability of an event failing is the complement of its success probability: P(E') = 1 - P(E).

2

Compute the probability that both switches fail simultaneously.

P(R_1' ∩ R_2') = 0.20 × 0.25 = 0.05

Since the switches operate independently, the multiplication law for independent events applies: P(A ∩ B) = P(A) × P(B).

3

Calculate the probability that at least one switch functions successfully.

P(at least one) = 1 - 0.05 = 0.95

The complement of 'neither switch functioning' is 'at least one switch functioning'.

Anahtar Kavram

Probability laws for independent compound events and the complement rule

Tahmini Süre:1m 30s

Soru 43Soru →

In how many different ways can $5$ boys and $3$ girls be seated in a straight row such that all $3$ girls must sit together?

Cevabı ve açıklamayı göster

Cevap: 4320

Cevap

The total number of ways to seat

5

boys and

3

girls in a row such that all

3

girls sit together is

4320

.

To arrange

5

boys and

3

girls so that the girls are always together, treat the

3

girls as

1

single unit. Combined with the

5

boys, there are

6

units to arrange in a straight line, which can be done in

6! = 720

ways. Within their group, the

3

girls can be arranged in

3! = 6

ways. By the multiplication principle, the total number of seating arrangements is

720 \times 6 = 4320

.

Adım Adım Çözüm

1

Group the restricted items into a single block

The

3

girls form

1

unit. Combined with the

5

boys, there are

5 + 1 = 6

units to arrange.

Since all

3

girls must sit together, treating them as a single block ensures they are not separated.

2

Calculate the linear arrangements of the combined units

The

6

units can be arranged in

6! = 720

ways.

The number of distinct ways to arrange

n

items in a line is

n!

.

3

Calculate internal arrangements of the girls' block

The

3

girls can be arranged among themselves in

3! = 6

ways.

The

3

girls inside the block are distinct individuals and can swap positions.

4

Apply the fundamental counting principle

Total arrangements =

6! \times 3! = 720 \times 6 = 4320

.

The total number of arrangements is the product of external block arrangements and internal block arrangements.

Anahtar Kavram

Permutations with grouping constraints (string method)

Soru 44Soru →

The mean of five numbers arranged in ascending order is $28$ . The mean of the first three numbers is $22$ , while the mean of the last three numbers is $36$ . Find the median of these five numbers.

Cevabı ve açıklamayı göster

Cevap: 34

Cevap

The median of the five numbers is

34

.

For five numbers ordered from smallest to largest (

x_1, x_2, x_3, x_4, x_5

), the median is the middle value

x_3

. The sum of all five numbers is

5 \times 28 = 140

. The sum of the first three numbers is

x_1 + x_2 + x_3 = 3 \times 22 = 66

, and the sum of the last three numbers is

x_3 + x_4 + x_5 = 3 \times 36 = 108

. Adding these two partial sums gives

(x_1 + x_2 + x_3 + x_4 + x_5) + x_3 = 66 + 108 = 174

. Substituting the overall sum of

140

into the equation yields

140 + x_3 = 174

, which simplifies to

x_3 = 34

.

Adım Adım Çözüm

1

Calculate the sum of all five numbers.

Sum of all 5 numbers is

5 \times 28 = 140

.

The total sum of a set of data equals the number of items multiplied by the mean.

2

Calculate the partial sums of the first three and last three numbers.

First three numbers sum to

3 \times 22 = 66

; last three numbers sum to

3 \times 36 = 108

.

Multiplying each sub-group mean by the count of numbers in that sub-group yields the sub-group sum.

3

Set up an equation relating the partial sums to the total sum and the median.

Adding the partial sums counts the third number (median) twice:

66 + 108 = 140 + \text{median}

.

In an ordered set of 5 numbers, the 3rd term is the median and is shared by both the first three and last three elements.

4

Solve for the median.

Median

= 174 - 140 = 34

.

Subtracting the total sum from the combined partial sums isolates the overlapping median value.

Anahtar Kavram

Measures of Central Tendency for Ungrouped Data (Relationship between sub-group means, total sum, and median in ordered data)

Soru 45Soru →

The masses, in kilograms, of five wooden crates stored in a warehouse are $3$ , $7$ , $8$ , $10$ , and $12$ . What is the variance of the masses of these crates?

9.2

2.4

11.5

46.0

Cevabı ve açıklamayı göster

Cevap: $9.2$

Cevap

The variance of the masses is

9.2

.

The mean of the numbers is

8

. The squared deviations are

25, 1, 0, 4,

and

16

. The sum of these squared deviations is

46

. Dividing by the count of numbers (

5

) gives

9.2

.

Adım Adım Çözüm

1

Calculate the mean (

\bar{x}

) of the dataset.

\bar{x} = \frac{3 + 7 + 8 + 10 + 12}{5} = \frac{40}{5} = 8

The mean is required to find individual deviations.

2

Compute the deviation of each observation from the mean,

(x - \bar{x})

.

3 - 8 = -5

,

7 - 8 = -1

,

8 - 8 = 0

,

10 - 8 = 2

,

12 - 8 = 4

Deviations quantify how far each data point lies from the central mean.

3

Square each deviation,

(x - \bar{x})^2

.

(-5)^2 = 25

,

(-1)^2 = 1

,

0^2 = 0

,

2^2 = 4

,

4^2 = 16

Squaring eliminates negative signs and weights larger deviations.

4

Sum the squared deviations and divide by the total number of observations

N = 5

.

Variance

\sigma^2 = \frac{25 + 1 + 0 + 4 + 16}{5} = \frac{46}{5} = 9.2

Variance is defined as the mean of the squared deviations.

Anahtar Kavram

Variance of Ungrouped Data

Tahmini Süre:1m 30s

Soru 46Soru →

A fruit vendor has $7$ distinct types of fresh fruits on display. A customer wants to buy a gift basket containing exactly $4$ different types of fruits. How many different combinations of fruits can the customer choose?

Cevabı ve açıklamayı göster

Cevap: 35

Cevap

The customer can choose the fruits in 35 different ways.

The number of ways to choose

r = 4

items from

n = 7

distinct items without regard to order is given by

^7C_4 = \frac{7!}{4!3!} = \frac{210}{6} = 35

.

Adım Adım Çözüm

1

Identify total elements (

n

) and selected subset size (

r

).

n = 7

and

r = 4

.

Since the selection order inside the fruit basket does not matter, combinations (

nCr

) must be used.

2

Substitute values into the combination formula

^nC_r = \frac{n!}{r!(n-r)!}

.

^7C_4 = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!}

.

This evaluates the total ways to choose 4 items from 7 without repetition or ordering.

3

Simplify the factorials and compute the numerical result.

\frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35

.

Canceling out

4!

from numerator and denominator gives

(7 \times 6 \times 5) / 6 = 35

.

Anahtar Kavram

Combinations (

nCr

)

Tahmini Süre:45s

Soru 47Soru →

The arithmetic mean of five observations $1$ , $3$ , $x$ , $9$ , and $12$ is $6$ . What is the variance of this set of observations?

3.6

4

16

20

Cevabı ve açıklamayı göster

Cevap: $16$

Cevap

The variance of the set of observations is

16

.

First, the missing value is found by equating the sum of the five numbers to

5 \times 6 = 30

, which gives

x = 5

. The data set is

1, 3, 5, 9, 12

. The deviations from the mean

6

are

-5, -3, -1, 3, 6

, and their squares are

25, 9, 1, 9, 36

. Summing these squared deviations gives

80

. Dividing by the number of observations (

5

) yields the variance of

16

.

Adım Adım Çözüm

1

Find the missing value

x

using the arithmetic mean formula.

x = 5

The mean of

N = 5

numbers is

\bar{x} = 6

, so the total sum is

5 \times 6 = 30

. Summing the known numbers gives

1 + 3 + x + 9 + 12 = 30 \implies 25 + x = 30 \implies x = 5

.

2

Calculate the deviations

(x_i - \bar{x})

for each data value from the mean

\bar{x} = 6

.

Deviations are

-5, -3, -1, 3, 6

Subtracting

6

from each value

(1, 3, 5, 9, 12)

gives

1-6=-5

,

3-6=-3

,

5-6=-1

,

9-6=3

, and

12-6=6

.

3

Square each deviation.

Squared deviations are

25, 9, 1, 9, 36

Squaring each deviation yields

(-5)^2 = 25

,

(-3)^2 = 9

,

(-1)^2 = 1

,

3^2 = 9

, and

6^2 = 36

.

4

Compute the population variance

\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{N}

.

\sigma^2 = 16

The sum of squared deviations is

25 + 9 + 1 + 9 + 36 = 80

. Dividing by

N = 5

gives

\frac{80}{5} = 16

.

Anahtar Kavram

Variance of Ungrouped Data with a Missing Observation

Soru 48Soru →

The daily temperature readings, in degrees Celsius, recorded over four consecutive days are $10$ , $12$ , $14$ , and $16$ . What is the mean deviation of these temperatures?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The mean deviation of the temperatures is 2 °C.

To find the mean deviation, first calculate the mean of the data values: (10 + 12 + 14 + 16) / 4 = 13. Next, find the absolute difference of each value from the mean: |10 - 13| = 3, |12 - 13| = 1, |14 - 13| = 1, and |16 - 13| = 3. Finally, average these absolute differences: (3 + 1 + 1 + 3) / 4 = 8 / 4 = 2.

Adım Adım Çözüm

1

Calculate the arithmetic mean (average) of the data set

$\bar{x} = \frac{10 + 12 + 14 + 16}{4} = 13$

Mean deviation measures dispersion relative to the mean, so the mean must be calculated first.

2

Compute the absolute deviation of each value from the mean

$|10 - 13| = 3$, $|12 - 13| = 1$, $|14 - 13| = 1$, $|16 - 13| = 3$

Mean deviation requires non-negative distances of each observation from the mean.

3

Sum the absolute deviations and divide by the sample size (N = 4)

$\text{Mean Deviation} = \frac{3 + 1 + 1 + 3}{4} = \frac{8}{4} = 2$

The mean deviation is the average of the absolute deviations.

Anahtar Kavram

Mean Deviation for Ungrouped Data

Soru 49Soru →

A football coach records the number of goals scored by a team in six consecutive matches as $1, 3, 4, 6, 7,$ and $9$ . What is the variance of the goals scored?

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The variance of the goals scored is 7.

To find the variance of the goals scored, first calculate the mean:

\bar{x} = \frac{1+3+4+6+7+9}{6} = \frac{30}{6} = 5

. Next, determine the sum of the squared deviations from the mean:

(1-5)^2 + (3-5)^2 + (4-5)^2 + (6-5)^2 + (7-5)^2 + (9-5)^2 = 16 + 4 + 1 + 1 + 4 + 16 = 42

. Dividing this total by the number of data values (

N = 6

) yields

\frac{42}{6} = 7

.

Adım Adım Çözüm

1

Calculate the mean of the dataset

\bar{x} = 5

The mean is needed as the reference point for computing deviations.

2

Compute the squared deviation of each data point from the mean

Squared deviations are 16, 4, 1, 1, 4, and 16

Variance measures the average squared distance from the mean.

3

Sum the squared deviations and divide by the number of observations N = 6

Variance = 7

The formula for variance of ungrouped data is \sigma^2 = \frac{\sum (x - \bar{x})^2}{N}.

Anahtar Kavram

Variance of Ungrouped Data

Soru 50Soru →

Calculate the total number of distinct four-digit numbers that can be formed using the digits $1, 2, 3, 4, 5, 6,$ and $7$ without repetition, such that the resulting number is divisible by either $4$ or $5$ .

Cevabı ve açıklamayı göster

Cevap: 320

Cevap

The total number of such four-digit numbers is

320

.

The total number of valid four-digit numbers is found by adding the count of numbers divisible by

5

(

120

) to the count of numbers divisible by

4

(

200

). Since a number ending in

5

is odd, it cannot be divisible by

4

, making the two conditions mutually exclusive. Thus, the total count is

120 + 200 = 320

.

Adım Adım Çözüm

1

Calculate the number of four-digit numbers divisible by

5

.

For a number to be divisible by

5

, its units digit must be

5

(since

0

is not available). There is

1

choice for the units digit. The remaining

3

positions are filled from the remaining

6

available digits in

^{6}P_{3} = 6 \times 5 \times 4 = 120

ways.

Divisibility by

5

requires the last digit to be

5

.

2

Calculate the number of four-digit numbers divisible by

4

.

A number is divisible by

4

if its last two digits form a multiple of

4

. Using distinct digits from

\{1, 2, 3, 4, 5, 6, 7\}

, the valid two-digit endings are

12, 16, 24, 32, 36, 52, 56, 64, 72,

and

76

(

10

valid pairs). For each pair, the first two positions are filled from the remaining

5

digits in

^{5}P_{2} = 5 \times 4 = 20

ways. Thus, total ways =

10 \times 20 = 200

.

Divisibility by

4

depends entirely on the last two digits.

3

Check for overlap (numbers divisible by both

4

and

5

).

A number divisible by

5

must end in

5

, which is an odd digit. All multiples of

4

must end in an even digit. Hence, no number is divisible by both

4

and

5

in this set. The overlap is

0

.

The two events are mutually exclusive.

4

Apply the addition principle of counting.

Total =

120 + 200 - 0 = 320

.

Add the counts of the two mutually exclusive sets.

Anahtar Kavram

Restricted Permutations and Mutually Exclusive Events

Tahmini Süre:1m 30s

Soru 51Soru →

The heights (in cm) of a seedling recorded over five consecutive weeks are $9$ , $13$ , $15$ , $17$ , and $21$ . What is the standard deviation of the heights?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The standard deviation of the seedling heights is 4 cm.

To find the standard deviation, first compute the mean: (9 + 13 + 15 + 17 + 21) / 5 = 15. Next, calculate the sum of squared deviations: (-6)^2 + (-2)^2 + 0^2 + 2^2 + 6^2 = 36 + 4 + 0 + 4 + 36 = 80. Divide by 5 to find the variance of 16. Finally, taking the square root of 16 gives the standard deviation of 4 cm.

Adım Adım Çözüm

1

Calculate the mean of the data values

Mean = 15 cm

The mean is necessary to calculate the deviations of each data point.

2

Calculate the squared deviations from the mean

Squared deviations are 36, 4, 0, 4, and 36 (sum = 80)

Variance measures the average of squared deviations from the mean.

3

Calculate the population variance

Variance = 80 / 5 = 16

Dividing the sum of squared deviations by N gives the variance.

4

Calculate the standard deviation

Standard deviation = sqrt(16) = 4 cm

The standard deviation is the square root of the variance.

Anahtar Kavram

Standard Deviation of Ungrouped Data

Soru 52Soru →

Find the number of distinct arrangements of the letters of the word $\text{PARALLEL}$ such that no two letters $\text{'L'}$ are adjacent.

Cevabı ve açıklamayı göster

Cevap: 1200

Cevap

1200

To ensure no two letters 'L' are adjacent, we use the gap method. First, arrange the 5 non-L letters (P, A, R, A, E). Because 'A' repeats twice, there are 5! / 2! = 60 distinct arrangements. These 5 letters form 6 available gaps (including the ends). Selecting 3 of these 6 gaps to insert the 3 identical 'L's can be done in C(6, 3) = 20 ways. Multiplying these gives 60 × 20 = 1200 valid arrangements.

Adım Adım Çözüm

1

Count the frequency of each letter in the word PARALLEL.

The word has 8 letters: 1 P, 2 A's, 1 R, 3 L's, and 1 E.

Recognizing repeated elements is essential for permutations with duplicates.

2

Arrange the non-restricted letters (P, A, R, A, E).

Number of arrangements = 5! / 2! = 60.

The letter 'A' is repeated twice, so we divide 5! by 2!.

3

Calculate the number of available gaps for placing the 3 'L's so that no two are adjacent.

5 arranged letters create 6 gap positions. Choosing 3 gaps gives C(6, 3) = (6 × 5 × 4) / (3 × 2 × 1) = 20 ways.

Placing at most one 'L' per gap guarantees that no two 'L's will be adjacent.

4

Multiply the number of arrangements of non-L letters by the gap choices.

Total arrangements = 60 × 20 = 1200.

By the fundamental counting principle, total arrangements equal the product of independent choices.

Anahtar Kavram

Permutations with repeated elements and non-adjacency constraints using the Gap Method

Tahmini Süre:1m 30s

Soru 53Soru →

A set of six numbers arranged in ascending order is given as $3, 8, x, 14, y, 25$ . If the median of the data set is $11$ and its arithmetic mean is $13$ , what is the value of $y$ ?

17

20

22

16

Cevabı ve açıklamayı göster

Cevap: $20$

Cevap

The value of

y

is

20

.

The median of an even number of values arranged in order is the average of the two central terms, which gives

\frac{x + 14}{2} = 11

, solving to

x = 8

. Since the mean of the six numbers is

13

, their total sum is

6 \times 13 = 78

. Adding all terms gives

3 + 8 + 8 + 14 + y + 25 = 58 + y = 78

, which yields

y = 20

.

Adım Adım Çözüm

1

Find the value of

x

using the median of the six ordered numbers.

\frac{x + 14}{2} = 11 \implies x + 14 = 22 \implies x = 8

For an even number of observations (

n = 6

), the median is the average of the 3rd term (

x

) and 4th term (

14

).

2

Calculate the required total sum of all six numbers using the given mean.

\text{Total Sum} = 6 \times 13 = 78

The mean of

n

numbers is equal to the sum of the numbers divided by

n

.

3

Sum all six terms and solve for

y

.

3 + 8 + 8 + 14 + y + 25 = 78 \implies 58 + y = 78 \implies y = 20

Substitute

x = 8

into the dataset and set the sum of all elements equal to

78

.

Anahtar Kavram

Measures of Central Tendency for Ungrouped Data

Soru 54Soru →

How many distinct five-digit even numbers greater than $50,000$ can be formed using the digits $2, 3, 5, 6, 7,$ and $8$ if no digit may be repeated in any number?

240

288

480

40

Cevabı ve açıklamayı göster

Cevap: 240

Cevap

The total number of distinct five-digit even numbers greater than 50,000 is 240.

The correct total is 240. Because the first digit requirement (greater than 50,000 requires 5, 6, 7, or 8) and the last digit requirement (even requires 2, 6, or 8) share common elements (6 and 8), the problem must be split into two mutually exclusive cases: when the first digit is odd (2 choices for

d_1

, 3 for

d_5

, and

^4P_3 = 24

for the middle digits, giving

2 \times 3 \times 24 = 144

) and when the first digit is even (2 choices for

d_1

, 2 for

d_5

, and

^4P_3 = 24

for the middle digits, giving

2 \times 2 \times 24 = 96

). Adding these yields

144 + 96 = 240

.

Adım Adım Çözüm

1

Identify constraints on the positions of a five-digit number

d_1 d_2 d_3 d_4 d_5

.

For the number to be greater than 50,000,

d_1 \in \{5, 6, 7, 8\}

. For the number to be even,

d_5 \in \{2, 6, 8\}

.

The first digit determines the magnitude and the last digit determines parity.

2

Analyze Case 1 where the first digit

d_1

is odd (

5

or

7

).

There are 2 choices for

d_1

. All 3 even digits remain available for

d_5

(3 choices). The remaining 3 middle positions (

d_2, d_3, d_4

) can be arranged from the 4 remaining unused digits in

^4P_3 = 24

ways. Total for Case 1 =

2 \times 3 \times 24 = 144

.

Choosing an odd first digit does not reduce the pool of available even digits for the units position.

3

Analyze Case 2 where the first digit

d_1

is even (

6

or

8

).

There are 2 choices for

d_1

. Since one even digit is used in

d_1

, 2 even digits remain for

d_5

(2 choices). The 3 middle positions are filled from the 4 remaining digits in

^4P_3 = 24

ways. Total for Case 2 =

2 \times 2 \times 24 = 96

.

Selecting an even digit for the ten-thousands place directly reduces the available choices for the units place.

4

Sum the possibilities from both mutually exclusive cases.

Total valid numbers =

144 + 96 = 240

.

Applying the addition principle of counting over disjoint cases.

Anahtar Kavram

Conditional permutations with non-disjoint position restrictions

Soru 55Soru →

The table below shows the frequency distribution of marks obtained by a group of students in a short test:

Mark ( $x$ )	1	2	3	4
Frequency ( $f$ )	2	3	3	2

What is the mean deviation of the distribution?

0.9

1.0

1.05

2.25

Cevabı ve açıklamayı göster

Cevap: $0.9$

Cevap

The mean deviation of the distribution is

0.9

.

The mean of the data is

\bar{x} = \frac{25}{10} = 2.5

. Calculating the sum of weighted absolute deviations yields

2(1.5) + 3(0.5) + 3(0.5) + 2(1.5) = 9.0

. Dividing by total frequency

10

gives the mean deviation as

0.9

.

Adım Adım Çözüm

1

Calculate the mean (

\bar{x}

) of the frequency distribution.

\sum f = 2 + 3 + 3 + 2 = 10

,

\sum fx = (2 \times 1) + (3 \times 2) + (3 \times 3) + (2 \times 4) = 2 + 6 + 9 + 8 = 25

. Therefore,

\bar{x} = \frac{25}{10} = 2.5

.

The mean is required to determine the deviations of each score value.

2

Find the absolute deviation

|x - \bar{x}|

for each score value.

For

x = 1

:

|1 - 2.5| = 1.5

; for

x = 2

:

|2 - 2.5| = 0.5

; for

x = 3

:

|3 - 2.5| = 0.5

; for

x = 4

:

|4 - 2.5| = 1.5

.

Mean deviation measures average distance from the mean, ignoring signs.

3

Multiply each absolute deviation by its corresponding frequency and sum them.

\sum f|x - \bar{x}| = (2 \times 1.5) + (3 \times 0.5) + (3 \times 0.5) + (2 \times 1.5) = 3 + 1.5 + 1.5 + 3 = 9.0

.

Frequencies reflect how many times each deviation occurs in the dataset.

4

Compute the mean deviation by dividing the weighted sum by the total frequency

\sum f

.

\text{Mean Deviation} = \frac{\sum f|x - \bar{x}|}{\sum f} = \frac{9.0}{10} = 0.9.

The mean deviation is the average of these absolute deviations across all observations.

Anahtar Kavram

Mean Deviation for Discrete Frequency Distribution

Soru 56Soru →

In a science competition, $3$ distinct prizes (first, second, and third place) are to be awarded to $3$ different students chosen from a group of $5$ finalists. In how many different ways can these $3$ prizes be awarded?

Cevabı ve açıklamayı göster

Cevap: 60

Cevap

60 ways

Because the prizes are distinct, the order in which the finalists are selected matters. Calculating the number of arrangements of

3

students from

5

candidates is given by

^{5}P_{3} = 5 \times 4 \times 3 = 60

.

Adım Adım Çözüm

1

Determine if order matters

Since the prizes are distinct (1st, 2nd, and 3rd place), the order of assignment matters, making this a permutation problem.

Assigning distinct positions to individuals requires calculating permutations rather than combinations.

2

Apply the permutation formula

^{n}P_{r} = \frac{n!}{(n-r)!}

^{5}P_{3} = \frac{5!}{(5-3)!} = \frac{5!}{2!}

There are

5

total candidates (

n = 5

) and

3

positions to fill (

r = 3

).

3

Calculate the numeric value

^{5}P_{3} = 5 \times 4 \times 3 = 60

Canceling

2!

from the numerator and denominator leaves

5 \times 4 \times 3 = 60

.

Anahtar Kavram

Linear permutation of r items selected from n distinct items

Tahmini Süre:45s

Soru 57Soru →

Match each statistical data representation term on the left with its corresponding definition or mathematical property on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Class Boundary matches with the value separating adjacent non-overlapping class intervals; Sector Angle matches with the central angle in a pie chart calculated as

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

; Frequency Density matches with the quotient of class frequency and class width; Ogive matches with a line graph produced by plotting cumulative frequencies against upper class boundaries.

Each data representation term directly corresponds to its core definition: class boundary closes gaps between discrete class intervals, sector angle measures central circle proportion in a pie chart, frequency density scales histogram height when class widths differ, and an ogive graphs cumulative frequency against upper boundaries.

Adım Adım Çözüm

1

Define Class Boundary

Class boundary is the continuous point midway between adjacent class limits.

Class boundaries remove gaps in discrete grouped frequency distributions.

2

Define Sector Angle formula for a pie chart

Sector Angle

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

.

The complete circle represents total frequency, so individual sectors scale proportionally with

360^\circ

.

3

Define Frequency Density for histograms

Frequency Density

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

.

Histogram area equals frequency; when widths differ, height must represent frequency per unit width.

4

Define Ogive

An Ogive is a cumulative frequency curve plotted against upper boundaries.

Each point on an ogive shows the cumulative frequency up to that class's upper boundary.

Anahtar Kavram

Data Representation Terms and Formulas

Tahmini Süre:1m 30s

Soru 58Soru →

The ages, in years, of five participants in a workshop are $2, 4, 5, 7,$ and $12$ . What is the mean deviation of the ages?

2.8

0.0

3.5

6.0

Cevabı ve açıklamayı göster

Cevap: $2.8$

Cevap

2.8

The mean deviation is calculated by taking the average of the absolute differences between each data point and the mean. For the dataset

2, 4, 5, 7,

and

12

, the mean is

6

. The absolute deviations from

6

are

4, 2, 1, 1,

and

6

, which sum to

14

. Dividing

14

by

5

gives

2.8

.

Adım Adım Çözüm

1

Calculate the arithmetic mean (

\bar{x}

) of the dataset.

\bar{x} = \frac{2 + 4 + 5 + 7 + 12}{5} = \frac{30}{5} = 6

The mean is needed as the reference point to calculate deviations.

2

Calculate the absolute deviation

|x - \bar{x}|

for each data item.

|2 - 6| = 4, \quad |4 - 6| = 2, \quad |5 - 6| = 1, \quad |7 - 6| = 1, \quad |12 - 6| = 6

Mean deviation measures distance from the mean, so absolute values are taken.

3

Find the average of these absolute deviations.

\text{Mean Deviation} = \frac{4 + 2 + 1 + 1 + 6}{5} = \frac{14}{5} = 2.8

The formula for mean deviation of ungrouped data is

\frac{\sum |x - \bar{x}|}{n}

.

Anahtar Kavram

Mean Deviation for Ungrouped Data

Soru 59Soru →

The frequency distribution table below shows the mass, in grams, of $20$ harvested oranges recorded during an agricultural experiment:

Mass ( $g$ )	Frequency ( $f$ )
$10 - 19$	$3$
$20 - 29$	$5$
$30 - 39$	$8$
$40 - 49$	$4$

What is the mean mass of the harvested oranges?

26.5\text{ g}

29.5\text{ g}

31.0\text{ g}

35.5\text{ g}

Cevabı ve açıklamayı göster

Cevap: $31.0\text{ g}$

Cevap

The mean mass of the harvested oranges is

31.0\text{ g}

.

The mean of grouped data is given by

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

. Finding the midpoints (

x

) of the intervals gives

14.5

,

24.5

,

34.5

, and

44.5

. Multiplying each midpoint by its frequency yields

43.5

,

122.5

,

276.0

, and

178.0

. The sum of these products is

620.0

, and dividing by the total frequency (

20

) gives

31.0\text{ g}

.

Adım Adım Çözüm

1

Determine the class midpoint (

x

) for each class interval.

Midpoints are:

\frac{10+19}{2} = 14.5

,

\frac{20+29}{2} = 24.5

,

\frac{30+39}{2} = 34.5

, and

\frac{40+49}{2} = 44.5

.

For grouped frequency data, each interval is represented by its central value (class mark or midpoint).

2

Multiply each midpoint (

x

) by its corresponding frequency (

f

) to calculate

fx

.

f_1 x_1 = 3 \times 14.5 = 43.5

,

f_2 x_2 = 5 \times 24.5 = 122.5

,

f_3 x_3 = 8 \times 34.5 = 276.0

,

f_4 x_4 = 4 \times 44.5 = 178.0

.

This computes the total mass contributed by the items in each interval.

3

Calculate the total sum of

fx

and total frequency

\sum f

.

\sum fx = 43.5 + 122.5 + 276.0 + 178.0 = 620.0

and

\sum f = 3 + 5 + 8 + 4 = 20

.

These sums are the numerator and denominator required for the mean formula.

4

Compute the mean using

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

.

\bar{x} = \frac{620.0}{20} = 31.0\text{ g}

.

Dividing the total weighted value by the total number of items gives the mean.

Anahtar Kavram

Calculating Mean for Grouped Frequency Data

Soru 60Soru →

A pie chart is used to display the distribution of $720$ candidates registered for a competitive examination across five subjects. If the sector representing Further Mathematics has a central angle of $45^\circ$ , what is the total number of candidates who registered for Further Mathematics?

Cevabı ve açıklamayı göster

Cevap: 90

Cevap

The total number of candidates who registered for Further Mathematics is 90.

To calculate the number of candidates represented by a pie chart sector, multiply the total count by the ratio of the sector's central angle to 360 degrees: (45 / 360) * 720 = 90 candidates.

Adım Adım Çözüm

1

Determine the fraction of the total population represented by the Further Mathematics sector.

\frac{45^\circ}{360^\circ} = \frac{1}{8}

A complete pie chart circle corresponds to an angle of 360 degrees.

2

Calculate the actual number of candidates by multiplying the fraction by the total student population.

\frac{1}{8} \times 720 = 90

The number of items in a sector is directly proportional to its central sector angle relative to 360 degrees.

Anahtar Kavram

Calculating category frequencies from pie chart sector angles

Statistics and Probability

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram