Statistics and Probability Alıştırma Soruları — JAMB UTME

The table below shows the distribution of quiz scores for a class of students:

Score Interval	Frequency ( $f$ )
$1 - 5$	$3$
$6 - 10$	$5$
$11 - 15$	$7$
$16 - 20$	$5$

Calculate the mean score of the distribution.

Cevabı ve açıklamayı göster

Cevap: 11.5

Cevap

The mean score of the distribution is 11.5.

The mean for a grouped frequency table is calculated by taking the sum of the products of each midpoint (

x

) and its frequency (

f

), divided by the sum of all frequencies (

f

). Here,

\sum fx = 230

and

\sum f = 20

, yielding

\bar{x} = 11.5

.

Adım Adım Çözüm

1

Determine the class midpoints (

x

) for each interval.

Midpoints are 3, 8, 13, and 18.

Midpoints serve as the representative values for each class interval.

2

Calculate the product of each midpoint and frequency (

fx

).

Products are 9, 40, 91, and 90.

To find the total contribution of each class interval.

3

Find the sum of all frequencies (

\sum f

) and products (

\sum fx

).

\sum f = 20

and

\sum fx = 230

.

These totals are required for the grouped mean formula.

4

Divide the total product sum by total frequency.

\bar{x} = \frac{230}{20} = 11.5

.

Applying the formula for the mean of grouped data

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

.

Anahtar Kavram

Mean of Grouped Data using Class Midpoints

Soru 2Soru →

Consider the following list of numbers representing test scores of seven students: $15$ , $4$ , $22$ , $9$ , $19$ , $2$ , and $13$ . What is the median of this data set?

9

12

13

15

Cevabı ve açıklamayı göster

Cevap: $13$

Cevap

The median of the data set is

13

.

First arrange the numbers in ascending order:

2, 4, 9, 13, 15, 19, 22

. Since there are

7

numbers, the median is the middle (

4\text{th}

) number, which is

13

.

Adım Adım Çözüm

1

Arrange the given numbers in ascending order.

The ordered data set is

2, 4, 9, 13, 15, 19, 22

.

To find the median of ungrouped data, the values must first be placed in numerical order.

2

Determine the position of the median for

n = 7

items.

Position =

\frac{n + 1}{2} = \frac{7 + 1}{2} = 4\text{th}

value.

For an odd number of observations

n

, the median is the

\frac{n+1}{2}\text{th}

term.

3

Identify the 4th value from the ordered list.

The 4th value is

13

.

Counting from the smallest value,

2

is 1st,

4

is 2nd,

9

is 3rd, and

13

is 4th.

Anahtar Kavram

Median of Ungrouped Data

Soru 3Soru →

The table below shows the distribution of weights (in kg) of 10 packages in a warehouse:

Weight Interval (kg)	Frequency ( $f$ )
$10 - 14$	$2$
$15 - 19$	$3$
$20 - 24$	$5$

What is the mean weight of the packages?

16.5\text{ kg}

17.0\text{ kg}

18.5\text{ kg}

20.5\text{ kg}

Cevabı ve açıklamayı göster

Cevap: $18.5\text{ kg}$

Cevap

The mean weight of the packages is

18.5\text{ kg}

.

The correct mean is calculated by finding the midpoint of each class interval (

12, 17, 22

), multiplying each by its frequency to obtain

fx

(

24, 51, 110

), and dividing the sum of

fx

(

185

) by the total frequency (

10

), giving

18.5\text{ kg}

.

Adım Adım Çözüm

1

Calculate the class midpoint (

x

) for each interval

Midpoints are:

\frac{10+14}{2} = 12

,

\frac{15+19}{2} = 17

, and

\frac{20+24}{2} = 22

.

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

2

Multiply each midpoint (

x

) by its corresponding frequency (

f

) to find

fx

2 \times 12 = 24

,

3 \times 17 = 51

, and

5 \times 22 = 110

.

This yields the total contribution of values from each class interval.

3

Sum the frequencies (

\sum f

) and the products (

\sum fx

)

\sum f = 2 + 3 + 5 = 10

and

\sum fx = 24 + 51 + 110 = 185

.

These sums are needed to compute the weighted mean.

4

Divide

\sum fx

by

\sum f

Mean

\bar{x} = \frac{185}{10} = 18.5\text{ kg}

.

The mean formula for grouped data is

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

.

Anahtar Kavram

Mean of Grouped Data using Class Marks

Soru 4Soru →

In how many different ways can a chairperson and a secretary be selected from a committee of $5$ members?

10

20

25

120

Cevabı ve açıklamayı göster

Cevap: $20$

Cevap

The number of ways to select a chairperson and a secretary from

5

members is

20

.

Selecting

2

individuals for distinct positions (chairperson and secretary) from

5

candidates is an ordered selection problem. The number of ways is given by the permutation formula

_5P_2 = 5 \times 4 = 20

.

Adım Adım Çözüm

1

Identify whether order matters

Order matters because the roles of chairperson and secretary are distinct.

When distinct roles are assigned, the arrangement is a permutation rather than a combination.

2

Apply the permutation formula

_nP_r = \frac{n!}{(n-r)!}

for

n=5

and

r=2

_5P_2 = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 5 \times 4 = 20

There are

5

choices for chairperson and

4

remaining choices for secretary.

Anahtar Kavram

Permutation of

n

distinct items taken

r

at a time

Tahmini Süre:45s

Soru 5Soru →

A standard six-sided die was rolled $100$ times during a probability experiment, and the outcome $4$ was recorded $25$ times. What is the experimental probability of rolling a $4$ ?

\frac{1}{4}

\frac{1}{6}

\frac{3}{4}

\frac{1}{25}

Cevabı ve açıklamayı göster

Cevap: $\frac{1}{4}$

Cevap

The experimental probability of rolling a

4

is

\frac{1}{4}

.

The experimental probability of an event is calculated by taking the ratio of the number of times the event occurs to the total number of trials performed. Since the outcome

4

appeared

25

times out of

100

rolls, the experimental probability is

\frac{25}{100}

, which simplifies directly to

\frac{1}{4}

.

Adım Adım Çözüm

1

Identify the number of favorable trials and total trials

Favorable trials (rolling a

4

) =

25

; Total trials =

100

.

Experimental probability depends on empirical data collected during the experiment.

2

Apply the experimental probability formula:

P(E) = \frac{\text{Frequency of Event}}{\text{Total Number of Trials}}

P(\text{rolling a } 4) = \frac{25}{100}

.

The experimental probability is defined as the relative frequency of the outcome.

3

Simplify the fraction to its lowest terms

\frac{25}{100} = \frac{1}{4}

.

Dividing both the numerator and denominator by their greatest common divisor,

25

, gives the simplified fraction.

Anahtar Kavram

Experimental Probability

Soru 6Soru →

In a mathematics test, the mean score of a group of $12$ boys is $60$ , and the mean score of a group of $18$ girls is $70$ . What is the mean score of all $30$ students combined?

66

65

64

67

Cevabı ve açıklamayı göster

Cevap: 66

Cevap

The combined mean score of all 30 students is 66.

To find the overall mean score for the entire class, determine the total sum of all test scores and divide by the total number of students. The total score contributed by the boys is

12 \times 60 = 720

, and the total score contributed by the girls is

18 \times 70 = 1260

. The combined total score is

720 + 1260 = 1980

. Dividing this total by

30

students yields a combined mean of

66

.

Adım Adım Çözüm

1

Calculate the total score for the boys

Total boys' score = 12 × 60 = 720

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

2

Calculate the total score for the girls

Total girls' score = 18 × 70 = 1260

Multiply the number of girls by their mean score.

3

Find the combined total score and total student count

Combined total score = 720 + 1260 = 1980; Total students = 12 + 18 = 30

Sum the total scores and the total counts for both groups.

4

Calculate the combined mean score

Combined mean = 1980 / 30 = 66

Divide the combined total score by the total number of students.

Anahtar Kavram

Combined Mean of Two Ungrouped Datasets

Soru 7Soru →

In a probability experiment, a card is drawn at random with replacement from a bag containing red, green, and blue cards. After conducting $250$ trials, a green card was drawn $85$ times. Given that the theoretical probability of drawing a green card is $0.30$ , calculate the positive difference between the experimental probability and the theoretical probability of drawing a green card.

Cevabı ve açıklamayı göster

Cevap: 0.04

Cevap

The positive difference between the experimental probability and the theoretical probability is 0.04.

The experimental probability is calculated as the ratio of observed favorable outcomes to total trials: $\frac{85}{250} = 0.34$. Subtracting the given theoretical probability of $0.30$ yields a positive difference of $|0.34 - 0.30| = 0.04$.

Adım Adım Çözüm

1

Determine experimental probability from trial data

Experimental probability = 85 / 250 = 0.34

Experimental probability is calculated as the ratio of observed favorable trials to the total number of trials executed.

2

Subtract theoretical probability from experimental probability

|0.34 - 0.30| = 0.04

Finding the positive difference requires subtracting the theoretical probability (0.30) from the experimental relative frequency (0.34).

Anahtar Kavram

Experimental probability is determined empirically by dividing the number of times an event occurs by the total number of trials, whereas theoretical probability is based on expected outcomes under ideal conditions.

Soru 8Soru →

In how many different ways can the letters of the word SUCCESS be arranged such that the three 'S's do not all come together?

360

420

60

35

Cevabı ve açıklamayı göster

Cevap: 360

Cevap

The letters of the word SUCCESS can be arranged in 360 ways such that the three 'S's do not all come together.

The correct answer is calculated using complementary counting. First, the total unrestricted permutations of SUCCESS (7 letters with 3 'S's and 2 'C's) is

\frac{7!}{3!2!} = 420

. Next, treating the three 'S's as one single block leaves 5 items to arrange with 2 'C's, giving

\frac{5!}{2!} = 60

ways where the 'S's are together. Subtracting 60 from 420 yields 360.

Adım Adım Çözüm

1

Calculate the total number of unrestricted arrangements of the word SUCCESS.

The word SUCCESS has 7 letters in total: 3 'S's, 2 'C's, 1 'U', and 1 'E'. Total arrangements

N_{total} = \frac{7!}{3! \times 2!} = \frac{5040}{6 \times 2} = 420

.

Repeated letters must be accounted for by dividing the factorial of the total count by the factorials of the counts of repeated letters.

2

Calculate the number of arrangements where the three 'S's are all together.

Treat the three 'S's as a single entity (SSS). We now arrange 5 entities: (SSS), U, C, C, E. Since 'C' appears twice,

N_{together} = \frac{5!}{2!} = \frac{120}{2} = 60

.

Grouping restricted identical items into a single block allows us to find the subset of arrangements where they stay together.

3

Subtract the number of 'together' arrangements from the total arrangements.

N_{not\_together} = N_{total} - N_{together} = 420 - 60 = 360

.

The complementary counting principle gives the number of ways where the restriction is satisfied.

Anahtar Kavram

Permutations with Repeated Elements and Complementary Counting

Tahmini Süre:1m 30s

Soru 9Soru →

Match each data representation concept on the left with its corresponding mathematical formula or definition on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Sector angle of a category in a pie chart matches

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

; Height of a histogram bar for equal class intervals matches Frequency of the class interval; Height of a histogram bar for unequal class intervals matches

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

; Upper class boundary of an interval

a - b

matches

b + 0.5

.

Each chart concept matches its standard definition: pie chart sector angles are proportional parts of

360^\circ

; histogram heights equal class frequencies when widths are equal, but equal frequency density when widths are unequal; and class boundaries adjust discrete limits by half the gap width.

Adım Adım Çözüm

1

Identify the formula for calculating pie chart sector angles.

Sector angle =

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

.

Pie charts distribute

360^\circ

proportionally according to the frequency of each category.

2

Determine histogram bar height representation under uniform class widths.

Bar height corresponds directly to the class frequency.

With uniform widths, the area of each rectangle is directly proportional to its height.

3

Determine histogram bar height representation under varying class widths.

Bar height corresponds to frequency density, calculated as

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

.

To maintain area proportional to frequency across varying widths, height must equal frequency divided by width.

4

Determine upper class boundary for discrete class intervals.

Upper boundary =

b + 0.5

.

Class boundaries eliminate gaps between discrete intervals by extending limits by half of the unit gap.

Anahtar Kavram

Data Representation Principles in Charts and Frequency Tables

Soru 10Soru →

The table below shows the score distribution of a group of students in a quiz:

Score ( $x$ )	1	2	3	4	5
Frequency ( $f$ )	4	3	$k$	2	1

If the mean score of the distribution is $2.5$ , what is the median score of the dataset?

2.5

3.0

2.0

4.0

Cevabı ve açıklamayı göster

Cevap: $2.5$

Cevap

The median score of the dataset is

2.5

.

To find the median, we first find the missing frequency

k

using the mean formula

\text{Mean} = \frac{\sum fx}{\sum f}

. Setting

\frac{23 + 3k}{10 + k} = 2.5

gives

k = 4

. With

k = 4

, the total number of observations is

N = 14

. The median is the average of the 7th and 8th scores in the ordered dataset. From cumulative frequencies, the 7th score is

2

and the 8th score is

3

. Thus, the median is

\frac{2 + 3}{2} = 2.5

.

Adım Adım Çözüm

1

Set up the equation for the mean to find the missing frequency

k

\text{Mean} = \frac{\sum fx}{\sum f} = \frac{1(4) + 2(3) + 3(k) + 4(2) + 5(1)}{4 + 3 + k + 2 + 1} = \frac{23 + 3k}{10 + k} = 2.5

The mean of an ungrouped frequency table is the sum of all values divided by total frequency.

2

Solve the linear equation for

k

23 + 3k = 2.5(10 + k) \implies 23 + 3k = 25 + 2.5k \implies 0.5k = 2 \implies k = 4

Cross-multiplying and grouping like terms isolates the variable

k

.

3

Determine the total frequency

N

and median positions

N = 4 + 3 + 4 + 2 + 1 = 14

. Median position = average of

\frac{14}{2}\text{th}

(

7\text{th}

) and

(\frac{14}{2} + 1)\text{th}

(

8\text{th}

) values.

For an even total number of observations

N

, the median is the arithmetic mean of the two central numbers.

4

Find the

7\text{th}

and

8\text{th}

scores using cumulative frequency and calculate median

Cumulative frequencies: Score 1 (1–4), Score 2 (5–7), Score 3 (8–11).

7\text{th}\text{ term} = 2

,

8\text{th}\text{ term} = 3

.

\text{Median} = \frac{2 + 3}{2} = 2.5

.

The 7th score is 2 and the 8th score is 3, making their average 2.5.

Anahtar Kavram

Calculating the median of ungrouped frequency data after finding a missing frequency using the mean

Tahmini Süre:2m 0s

Soru 11Soru →

The table below shows the distribution of masses (in kg) of cocoa bags harvested on a farm:

Mass (kg)	Frequency ( $f$ )
$10 - 19$	$6$
$20 - 29$	$k$
$30 - 39$	$15$
$40 - 49$	$12$
$50 - 59$	$7$

If the estimated mean mass of the distribution is $35.3\text{ kg}$ , calculate the value of the missing frequency $k$ .

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

The value of the missing frequency is 10.

To find the missing frequency, compute the midpoints (

x

) of each mass class interval: 14.5, 24.5, 34.5, 44.5, and 54.5. Next, express the sum of frequencies as

\sum f = 40 + k

and the sum of products of frequency and midpoint as

\sum fx = 6(14.5) + k(24.5) + 15(34.5) + 12(44.5) + 7(54.5) = 1520 + 24.5k

. Equating the mean expression

\frac{\sum fx}{\sum f}

to

35.3

gives

\frac{1520 + 24.5k}{40 + k} = 35.3

. Cross-multiplying and solving yields

1520 + 24.5k = 1412 + 35.3k

, which simplifies to

10.8k = 108

, giving

k = 10

.

Adım Adım Çözüm

1

Determine the class midpoints (

x

) for all intervals.

Class midpoints are 14.5, 24.5, 34.5, 44.5, and 54.5.

Grouped mean calculations require representative midpoint values for each class interval.

2

Formulate expressions for total frequency

\sum f

and total weighted sum

\sum fx

.

\sum f = 40 + k and \sum fx = 1520 + 24.5k.

These algebraic expressions are necessary to substitute into the mean formula.

3

Set up and solve the linear equation using the given mean of 35.3.

\frac{1520 + 24.5k}{40 + k} = 35.3 \implies 10.8k = 108 \implies k = 10.

Equating the algebraic mean expression to the numerical mean allows solving for the unknown frequency k.

Anahtar Kavram

Measures of Central Tendency for Grouped Data

Soru 12Soru →

What is the mean deviation of the data set $4, 7, 8, 11, 15$ ?

Cevabı ve açıklamayı göster

Cevap: 3.2

Cevap

The mean deviation of the data set is 3.2.

To find the mean deviation, first calculate the mean of the dataset, which is 9. Then, compute the absolute difference of each number from 9, obtaining values of 5, 2, 1, 2, and 6. Finally, divide the sum of these absolute values (16) by the total number of items (5) to get 3.2.

Adım Adım Çözüm

1

Calculate the arithmetic mean of the given data set.

The mean

\bar{x} = 9

.

The mean is needed to evaluate how far each data point deviates from the central value.

2

Find the absolute difference between each value and the mean.

The absolute deviations are 5, 2, 1, 2, and 6.

Mean deviation measures dispersion using absolute distances, ignoring negative signs.

3

Calculate the mean of the absolute deviations.

Mean deviation = 3.2.

Dividing the total sum of absolute deviations (16) by the number of observations (5) gives the mean deviation.

Anahtar Kavram

Mean Deviation for Ungrouped Data

Soru 13Soru →

The table below shows the distribution of scores obtained by $60$ candidates in a competitive aptitude test:

Score Class Interval	Frequency ( $f$ )
$10 - 19$	$5$
$20 - 29$	$10$
$30 - 39$	$18$
$40 - 49$	$15$
$50 - 59$	$12$

Using cumulative frequency estimation (or linear interpolation), what is the interquartile range of the distribution?

18.0

20.0

30.0

15.0

Cevabı ve açıklamayı göster

Cevap: $18.0$

Cevap

The interquartile range of the distribution is

18.0

.

The lower quartile (

Q_1

) corresponds to the

15^{\text{th}}

rank, which is exactly

29.5

. The upper quartile (

Q_3

) corresponds to the

45^{\text{th}}

rank, which interpolates to

47.5

. Subtracting

Q_1

from

Q_3

yields

47.5 - 29.5 = 18.0

.

Adım Adım Çözüm

1

Construct the cumulative frequency distribution table with upper class boundaries.

Class intervals, upper class boundaries (

UCB

), frequencies (

f

), and cumulative frequencies (

cf

):
-

10 - 19

:

UCB = 19.5

,

f = 5

,

cf = 5

-

20 - 29

:

UCB = 29.5

,

f = 10

,

cf = 15

-

30 - 39

:

UCB = 39.5

,

f = 18

,

cf = 33

-

40 - 49

:

UCB = 49.5

,

f = 15

,

cf = 48

-

50 - 59

:

UCB = 59.5

,

f = 12

,

cf = 60

Cumulative frequencies and upper class boundaries are necessary to determine quartile ranks and values.

2

Find the lower quartile (

Q_1

).

The position of

Q_1

is

\frac{1}{4} N = \frac{1}{4}(60) = 15^{\text{th}}

candidate score. Since the cumulative frequency reaching

UCB = 29.5

is exactly

15

,

Q_1 = 29.5

.

The 15th value lies precisely at the upper class boundary of the

20 - 29

class interval.

3

Find the upper quartile (

Q_3

) using linear interpolation.

The position of

Q_3

is

\frac{3}{4} N = \frac{3}{4}(60) = 45^{\text{th}}

score.
This falls in the

40 - 49

class interval (

UCB = 49.5

, lower boundary

L = 39.5

,

f = 15

, previous

cf = 33

).

Q_3 = L + \left(\frac{\frac{3N}{4} - cf_{\text{prev}}}{f}\right) \times c = 39.5 + \left(\frac{45 - 33}{15}\right) \times 10 = 39.5 + 8.0 = 47.5

.

Linear interpolation calculates the exact position within the target class interval.

4

Calculate the Interquartile Range (IQR).

\text{IQR} = Q_3 - Q_1 = 47.5 - 29.5 = 18.0

.

The interquartile range is defined as the difference between the upper quartile and the lower quartile.

Anahtar Kavram

Interquartile Range estimation from Ogive / Cumulative Frequency Distribution

Tahmini Süre:1m 30s

Soru 14Soru →

The mean of eight consecutive odd numbers is $24$ . What is the median of the first four numbers?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

The median of the first four numbers is

20

.

Letting the eight consecutive odd numbers be

x, x+2, \dots, x+14

, their sum is

8x + 56

. Dividing by

8

gives a mean of

x + 7 = 24

, so

x = 17

. The first four numbers are

17, 19, 21,

and

23

. The median of these four values is the average of the middle two values (

19

and

21

), which equals

20

.

Adım Adım Çözüm

1

Represent the eight consecutive odd numbers algebraically

Let the numbers be

x, x+2, x+4, x+6, x+8, x+10, x+12, x+14

.

Consecutive odd numbers increase by steps of

2

.

2

Set up and solve the mean equation

\frac{(x) + (x+2) + (x+4) + (x+6) + (x+8) + (x+10) + (x+12) + (x+14)}{8} = 24 \implies x + 7 = 24 \implies x = 17

.

The mean of ungrouped data is the sum of all data values divided by the total number of values.

3

Identify the first four numbers in the set

The first four numbers are

17, 19, 21, 23

.

Substitute

x = 17

into

x, x+2, x+4,

and

x+6

.

4

Find the median of the first four numbers

\text{Median} = \frac{19 + 21}{2} = 20

.

For an even count of ordered values (4 items), the median is the arithmetic mean of the two middle terms.

Anahtar Kavram

Measures of Central Tendency for Ungrouped Data

Soru 15Soru →

A committee of $3$ members is to be selected from a group of $7$ people. In how many different ways can this committee be formed?

Cevabı ve açıklamayı göster

Cevap: 35

Cevap

35

Selecting a committee of

3

members from

7

people requires calculating the number of combinations, given by

^7C_3 = \frac{7!}{3!4!} = 35

.

Adım Adım Çözüm

1

Identify total elements and group size

n = 7

and

r = 3

Since the arrangement or order of members in the committee does not matter, this is a selection problem (combinations).

2

Apply the combinations formula

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

^7C_3 = \frac{7!}{3!4!}

This formula counts the distinct subsets of size

r

that can be chosen from

n

items.

3

Evaluate the factorial expression

^7C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35

Expanding

7!

as

7 \times 6 \times 5 \times 4!

allows cancelling

4!

, leaving

\frac{210}{6} = 35

.

Anahtar Kavram

Combinations formula

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

Tahmini Süre:45s

Soru 16Soru →

The set of numbers $k - 2$ , $k$ , $k + 1$ , and $k + 5$ is given, where $k$ is any real constant. What is the variance of this set of numbers?

Cevabı ve açıklamayı göster

Cevap: 6.5

Cevap

The variance of the given set of numbers is 6.5.

The mean of the set is

\bar{x} = k + 1

. Subtracting the mean from each data point gives deviations of

-3

,

-1

,

0

, and

4

. The squares of these deviations are

9

,

1

,

0

, and

16

, which sum to

26

. Dividing this sum by

4

gives a variance of

6.5

. A key statistical property illustrated here is that adding or subtracting a constant

k

from every value in a dataset shifts the mean by

k

but leaves measures of dispersion (such as variance and standard deviation) unchanged.

Adım Adım Çözüm

1

Find the mean (\bar{x}) of the given set {k - 2, k, k + 1, k + 5}.

\bar{x} = \frac{(k - 2) + k + (k + 1) + (k + 5)}{4} = \frac{4k + 4}{4} = k + 1

The mean is calculated by summing all values and dividing by the total count of numbers.

2

Determine the deviation of each value from the mean, (x_i - \bar{x}).

(k - 2) - (k + 1) = -3, k - (k + 1) = -1, (k + 1) - (k + 1) = 0, (k + 5) - (k + 1) = 4

Deviations measure how far each data value lies from the mean.

3

Square each individual deviation and sum the results.

(-3)^2 + (-1)^2 + 0^2 + 4^2 = 9 + 1 + 0 + 16 = 26

Squaring converts all deviations into non-negative values.

4

Divide the sum of squared deviations by the total number of observations (N = 4) to find the variance.

\text{Variance} = \frac{26}{4} = 6.5

Variance is defined as the arithmetic mean of the squared deviations from the mean.

Anahtar Kavram

Variance and Invariance under Constant Translation

Soru 17Soru →

Two independent events, $A$ and $B$ , have probabilities $P(A) = 0.4$ and $P(B) = 0.5$ . What is the probability that both event $A$ and event $B$ occur, $P(A \cap B)$ ?

Cevabı ve açıklamayı göster

Cevap: 0.2

Cevap

The probability that both events occur is 0.2.

For independent events, the joint probability of both events occurring simultaneously is found by multiplying their individual probabilities:

P(A \cap B) = P(A) \times P(B) = 0.4 \times 0.5 = 0.2

.

Adım Adım Çözüm

1

Identify event independence and the required probability operation

Events

A

and

B

are independent, and the question requires calculating their intersection

P(A \cap B)

.

The problem explicitly states that the events are independent.

2

Apply the multiplication law for independent events

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

For any two independent events, the probability of both occurring together is the product of their individual probabilities.

3

Perform the multiplication

P(A \cap B) = 0.4 \times 0.5 = 0.2

Multiplying 0.4 by 0.5 gives 0.2.

Anahtar Kavram

Multiplication Law of Probability for Independent Events

Soru 18Soru →

The mean mark of a student in $6$ tests is $14$ . When the highest and lowest marks, which differ by $12$ , are excluded, the mean mark of the remaining $4$ tests becomes $13.5$ . What is the highest mark?

Cevabı ve açıklamayı göster

Cevap: 21

Cevap

The highest mark is 21.

The total sum of all 6 tests is

6 \times 14 = 84

. When the highest (

H

) and lowest (

L

) marks are removed, the total sum of the remaining 4 tests is

4 \times 13.5 = 54

. The sum of the excluded marks is

H + L = 84 - 54 = 30

. Knowing that their difference is

H - L = 12

, we add the two equations to get

2H = 42

, which gives

H = 21

.

Adım Adım Çözüm

1

Calculate the total sum of all 6 test marks

Sum of 6 marks =

6 \times 14 = 84

The sum of data values equals the mean multiplied by the number of items.

2

Calculate the sum of the remaining 4 test marks

Sum of 4 marks =

4 \times 13.5 = 54

Multiplying the new mean by 4 gives the sum of the test scores excluding the highest and lowest values.

3

Determine the combined sum of the highest (H) and lowest (L) marks

H + L = 84 - 54 = 30

Subtracting the sum of the 4 remaining marks from the total initial sum yields the sum of the two excluded marks.

4

Solve for H using the simultaneous linear equations

H = 21

Adding H + L = 30 and H - L = 12 gives 2H = 42, which solves to H = 21.

Anahtar Kavram

Mean of Ungrouped Data and Handling Excluded Values

Soru 19Soru →

A convex polygon has $54$ diagonals. How many distinct triangles can be formed by joining any three of its vertices?

Cevabı ve açıklamayı göster

Cevap: 220

Cevap

220 distinct triangles

Solving the equation for the number of diagonals

\frac{n(n-3)}{2} = 54

yields

n = 12

vertices. The number of triangles that can be formed by selecting any 3 of these 12 vertices is given by

\binom{12}{3} = \frac{12 \times 11 \times 10}{6} = 220

.

Adım Adım Çözüm

1

Determine the number of vertices

n

of the polygon using the diagonals formula.

n = 12

The number of diagonals

D

in an

n

-sided convex polygon is given by

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

. Setting

\frac{n(n-3)}{2} = 54

gives

n^2 - 3n - 108 = 0

. Factoring

(n - 12)(n + 9) = 0

yields

n = 12

since the number of vertices must be positive.

2

Calculate the number of distinct triangles formed by choosing 3 vertices from 12.

220

Each set of 3 distinct vertices forms one unique triangle. The order in which the vertices are chosen does not matter, so we use combinations:

\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220

.

Anahtar Kavram

Combinations applied to geometric figures and polygon properties

Tahmini Süre:2m 0s

Soru 20Soru →

A biased four-sector spinner has sectors labeled 1, 2, 3, and 4. The theoretical probabilities of landing on sectors 1, 2, and 3 are in the ratio $1 : 2 : 2$ , respectively, while the theoretical probability of landing on sector 4 is $0.20$ . In an experiment consisting of $200$ spins, sector 2 was recorded $80$ times and sector 4 was recorded $60$ times. What is the positive difference between the expected number of even outcomes in $500$ future spins based on the experimental probability and that based on the theoretical probability?

90

50

318

60

Cevabı ve açıklamayı göster

Cevap: $90$

Cevap

90

The theoretical probability of getting an even outcome (sector 2 or 4) is

P(2) + P(4) = 0.32 + 0.20 = 0.52

, which predicts

500 \times 0.52 = 260

even outcomes in

500

spins. Empirically, even outcomes occurred

80 + 60 = 140

times in

200

spins, giving an experimental probability of

140 / 200 = 0.70

and an estimated frequency of

500 \times 0.70 = 350

. The positive difference between these two expected values is

350 - 260 = 90

.

Adım Adım Çözüm

1

Calculate the theoretical probabilities for each sector

P(1) = 0.16

,

P(2) = 0.32

,

P(3) = 0.32

,

P(4) = 0.20

Since

P(4) = 0.20

, the remaining probability

1 - 0.20 = 0.80

is shared among sectors 1, 2, and 3 in the ratio

1 : 2 : 2

(total 5 parts). Each part equals

0.80 / 5 = 0.16

.

2

Determine the theoretical probability and expected frequency of an even outcome

P_{\text{theo}}(\text{Even}) = 0.52

; Expected frequency in 500 spins

= 260

The even outcomes are sectors 2 and 4.

P_{\text{theo}}(\text{Even}) = P(2) + P(4) = 0.32 + 0.20 = 0.52

. In

500

spins,

500 \times 0.52 = 260

.

3

Calculate the experimental probability and expected frequency of an even outcome

P_{\text{exp}}(\text{Even}) = 0.70

; Expected frequency in 500 spins

= 350

Out of

200

spins, even outcomes occurred

80 + 60 = 140

times.

P_{\text{exp}}(\text{Even}) = 140 / 200 = 0.70

. In

500

spins,

500 \times 0.70 = 350

.

4

Find the positive difference between the two expected values

350 - 260 = 90

Subtracting the theoretical expected frequency from the experimental expected frequency gives

|350 - 260| = 90

.

Anahtar Kavram

Experimental vs. Theoretical Probability and Expected Frequency

Statistics and Probability

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