Statistics and Probability Alıştırma Soruları — JAMB UTME — Sayfa 5

The frequency distribution table below shows the scores obtained by $50$ candidates in a competitive assessment:

Score Interval	Frequency
$10 - 19$	$5$
$20 - 29$	$12$
$30 - 39$	$18$
$40 - 49$	$10$
$50 - 59$	$5$

When constructing a cumulative frequency curve (ogive) for this distribution, which of the following ordered pairs $(x, y)$ represents the point plotted for the class interval $30 - 39$ ?

(29.5, 35)

(39.5, 35)

(34.5, 35)

(39.5, 18)

Cevabı ve açıklamayı göster

Cevap: $(39.5, 35)$

Cevap

The point plotted on the ogive for the class interval

30 - 39

is

(39.5, 35)

.

A cumulative frequency curve (ogive) is constructed by plotting the cumulative frequency of each class against its upper class boundary. For the interval

30 - 39

, the upper boundary is

39.5

and the cumulative frequency is

5 + 12 + 18 = 35

, resulting in the point

(39.5, 35)

.

Adım Adım Çözüm

1

Determine the upper class boundary for the class interval

30 - 39

Upper class boundary =

\frac{39 + 40}{2} = 39.5

Cumulative frequencies in an ogive are plotted against the upper boundaries of each class interval.

2

Calculate the cumulative frequency up to the class interval

30 - 39

Cumulative frequency =

5 + 12 + 18 = 35

The cumulative frequency is the running sum of frequencies up to and including the given interval.

3

Combine the upper class boundary (

x

-axis) and cumulative frequency (

y

-axis) into an ordered pair

(x, y)

Point =

(39.5, 35)

In an ogive, points are plotted as (Upper Class Boundary, Cumulative Frequency).

Anahtar Kavram

Plotting Points on a Cumulative Frequency Curve (Ogive)

Soru 82Soru →

A meteorologist recorded the rainfall (in millimeters) in a town over five days as $2$ , $5$ , $6$ , $7$ , and $10$ . What is the mean deviation of the rainfall data?

2

0

10

6

Cevabı ve açıklamayı göster

Cevap: $2$

Cevap

The mean deviation of the rainfall data is

2

.

To find the mean deviation, first calculate the mean of the data:

(2 + 5 + 6 + 7 + 10) / 5 = 6

. Next, calculate the absolute difference of each value from 6, which yields 4, 1, 0, 1, and 4. The average of these absolute differences is

(4 + 1 + 0 + 1 + 4) / 5 = 10 / 5 = 2

.

Adım Adım Çözüm

1

Calculate the arithmetic mean (

\bar{x}

) of the dataset.

\bar{x} = \frac{2 + 5 + 6 + 7 + 10}{5} = \frac{30}{5} = 6

The mean is required to compute the deviation of each observation from the central value.

2

Find the absolute deviation

|x - \bar{x}|

for each data point.

|2 - 6| = 4, \quad |5 - 6| = 1, \quad |6 - 6| = 0, \quad |7 - 6| = 1, \quad |10 - 6| = 4

Mean deviation measures the average absolute distance of data values from the mean.

3

Sum the absolute deviations and divide by the total number of observations (

N = 5

).

\text{Mean Deviation} = \frac{4 + 1 + 0 + 1 + 4}{5} = \frac{10}{5} = 2

Dividing the sum of absolute deviations by

N

gives the mean deviation.

Anahtar Kavram

Mean Deviation

Soru 83Soru →

In how many distinct ways can a president, a vice-president, and a secretary be chosen from a group of $7$ candidates, assuming no candidate can hold more than one position?

Cevabı ve açıklamayı göster

Cevap: 210

Cevap

The total number of distinct ways to choose the three officers from 7 candidates is 210.

Selecting 3 distinct officers from a group of 7 candidates requires ordering 3 individuals out of 7, which equals

7 \times 6 \times 5 = 210

ways.

Adım Adım Çözüm

1

Identify whether the problem involves permutations or combinations

Order is important because the positions (President, Vice-President, Secretary) are distinct.

Selecting person X as President and person Y as Secretary is different from selecting person Y as President and person X as Secretary.

2

Calculate the permutation

^7P_3

^7P_3 = 7 \times 6 \times 5 = 210

There are 7 choices for President, 6 remaining choices for Vice-President, and 5 remaining choices for Secretary.

Anahtar Kavram

Permutations of n items taken r at a time

Tahmini Süre:45s

Soru 84Soru →

In how many distinct ways can the letters of the word $\text{KADUNA}$ be arranged?

720

15

360

180

Cevabı ve açıklamayı göster

Cevap: $360$

Cevap

The total number of distinct arrangements is

360

.

The word

\text{KADUNA}

consists of

6

total letters with the letter '\text{A}' repeated

2

times. Using the formula for permutations with identical items, the number of distinct arrangements is

\frac{6!}{2!} = \frac{720}{2} = 360

.

Adım Adım Çözüm

1

Count the total number of letters in the word

\text{KADUNA}

and identify repetitions.

Total letters

n = 6

. The letter '\text{A}' appears

2

times, while '\text{K}', '\text{D}', '\text{U}', and '\text{N}' each appear

1

time.

Arrangements of

n

items with repeated elements require dividing

n!

by the factorial of the count of each repeated element.

2

Apply the permutation formula for repeated elements:

P = \frac{n!}{p!}

where

p

is the frequency of the repeated letter.

P = \frac{6!}{2!} = \frac{720}{2} = 360

Dividing by

2!

eliminates duplicate arrangements caused by swapping identical letters.

Anahtar Kavram

Permutations of items with repeated elements

Tahmini Süre:45s

Soru 85Soru →

The mean of five numbers $3, 5, 7, x,$ and $y$ is $6$ . Given that the variance of the numbers is $8$ and $x < y$ , calculate the value of $y$ .

Cevabı ve açıklamayı göster

Cevap: 11

Cevap

11

The total sum of the five numbers is

5 \times 6 = 30

, giving

x + y = 15

. The sum of squared deviations from the mean

6

is

5 \times 8 = 40

. The known numbers

3, 5, 7

contribute

(-3)^2 + (-1)^2 + 1^2 = 11

to this sum, leaving

(x-6)^2 + (y-6)^2 = 29

. Substituting

y = 15 - x

yields

2x^2 - 30x + 88 = 0 \implies x^2 - 15x + 44 = 0

. Factoring gives roots

4

and

11

. Since

x < y

, we find

y = 11

.

Adım Adım Çözüm

1

Use the definition of the arithmetic mean to write a linear relationship between

x

and

y

.

x + y = 15

, or

y = 15 - x

.

The total sum of 5 numbers with a mean of 6 is

5 \times 6 = 30

. Subtracting the known numbers

3 + 5 + 7 = 15

leaves

x + y = 15

.

2

Apply the variance formula for population data.

\sum (x_i - 6)^2 = 40

.

Variance is the mean of squared deviations from the mean:

\frac{\sum (x_i - \bar{x})^2}{5} = 8 \implies \sum (x_i - 6)^2 = 40

.

3

Compute the sum of squared deviations for the known elements and simplify the variance equation.

(x - 6)^2 + (y - 6)^2 = 29

.

The squared deviations for

3, 5, 7

are

(-3)^2 = 9

,

(-1)^2 = 1

, and

1^2 = 1

. Subtracting

9 + 1 + 1 = 11

from

40

gives

29

.

4

Substitute

y = 15 - x

into the simplified equation and solve the resulting quadratic equation.

x = 4

or

x = 11

.

Substituting

y = 15 - x

gives

(x - 6)^2 + (9 - x)^2 = 29 \implies 2x^2 - 30x + 88 = 0 \implies x^2 - 15x + 44 = 0 \implies (x - 4)(x - 11) = 0

.

5

Select the correct pair

(x, y)

using the condition

x < y

.

x = 4

and

y = 11

.

Since

x < y

,

x

must be the smaller value (

4

) and

y

must be the larger value (

11

).

Anahtar Kavram

Calculation of variance and mean for ungrouped data containing unknown elements

Soru 86Soru →

The frequency distribution table below shows the marks obtained by $40$ students in a mathematics quiz:

Mark Interval	Frequency ( $f$ )
$1 - 10$	$5$
$11 - 20$	$8$
$21 - 30$	$12$
$31 - 40$	$10$
$41 - 50$	$5$

From a cumulative frequency curve (ogive) constructed for this data, what is the score corresponding to the $40^{\text{th}}$ percentile?

23.0

23.5

25.5

28.0

Cevabı ve açıklamayı göster

Cevap: $23.0$

Cevap

23.0

To find the 40th percentile score, calculate

40\%

of the total

40

students, which gives the

16^{\text{th}}

cumulative student. The

16^{\text{th}}

student falls in the

21 - 30

mark interval, which has a lower class boundary of

20.5

. Interpolating gives

20.5 + \left(\frac{16 - 13}{12}\right) \times 10 = 20.5 + 2.5 = 23.0

.

Adım Adım Çözüm

1

Construct the cumulative frequency distribution table.

Cumulative frequencies (

CF

):

1-10 \rightarrow 5

;

11-20 \rightarrow 13

;

21-30 \rightarrow 25

;

31-40 \rightarrow 35

;

41-50 \rightarrow 40

. Total frequency

N = 40

.

Cumulative frequencies are needed to locate percentile positions on an ogive.

2

Determine the rank position for the 40th percentile (

P_{40}

).

Rank position

= 0.40 \times 40 = 16^{\text{th}}

position.

The 40th percentile represents the value below which

40\%

of the observations fall.

3

Identify the percentile class interval and its boundaries.

Since

13 < 16 \leq 25

, the

16^{\text{th}}

position lies in the interval

21 - 30

. Lower class boundary

L = 20.5

, cumulative frequency preceding the class

CF_b = 13

, class frequency

f = 12

, and class width

c = 10

.

Interpolation on an ogive relies on the lower class boundary of the target interval.

4

Apply the linear interpolation formula for percentiles.

P_{40} = L + \left( \frac{\text{Rank} - CF_b}{f} \right) \times c = 20.5 + \left( \frac{16 - 13}{12} \right) \times 10 = 20.5 + 2.5 = 23.0

.

This yields the exact mark corresponding to the 40th percentile reading from the ogive.

Anahtar Kavram

Calculating percentiles from a cumulative frequency distribution or ogive

Tahmini Süre:1m 30s

Soru 87Soru →

The operating lifespans (in hours) of a sample of $60$ newly manufactured micro-components tested under laboratory conditions are presented in the frequency table below:

Lifespan (hours)	Frequency ( $f$ )
$10 - 19$	$5$
$20 - 29$	$8$
$30 - 39$	$12$
$40 - 49$	$20$
$50 - 59$	$15$

Using linear interpolation on class boundaries, calculate the median lifespan of the components in hours.

Cevabı ve açıklamayı göster

Cevap: 42

Cevap

The median lifespan of the micro-components is 42 hours.

To calculate the median of grouped data, first determine cumulative frequencies: 5, 13, 25, 45, 60. Total frequency is

N = 60

, placing the median at

N/2 = 30

. The median class is

40 - 49

with lower class boundary

L = 39.5

, class frequency

f = 20

, cumulative frequency prior to median class

c.f. = 25

, and class width

c = 10

. Substituting into

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

gives

39.5 + \left(\frac{30 - 25}{20}\right) \times 10 = 39.5 + 2.5 = 42

.

Adım Adım Çözüm

1

Calculate the total frequency

N

and the median rank

N/2

.

N = 5 + 8 + 12 + 20 + 15 = 60

, so

N/2 = 30

.

The median corresponds to the value at position 30 in the ordered cumulative frequency distribution.

2

Determine cumulative frequencies to locate the median class.

Cumulative frequencies are 5, 13, 25, 45, 60. The 30th value falls in the interval

40 - 49

.

The cumulative frequency first reaches or exceeds 30 at the

40 - 49

class.

3

Identify boundary parameters for the median class.

Lower boundary

L = 39.5

, class width

c = 10

, class frequency

f = 20

, and cumulative frequency of preceding class

c.f. = 25

.

Continuous data analysis requires using lower class boundaries rather than class limits.

4

Apply the grouped median linear interpolation formula.

\text{Median} = 39.5 + \left(\frac{30 - 25}{20}\right) \times 10 = 39.5 + 2.5 = 42

.

Linear interpolation within the median class yields the precise median value.

Anahtar Kavram

Grouped Data Median via Class Boundary Interpolation

Tahmini Süre:2m 0s

Soru 88Soru →

Five numbers $3, 7, 8, x,$ and $12$ are arranged in ascending order, where $x > 8$ . If the mean of the data set is equal to its median, what is the value of $x$ ?

10

5

14

18

Cevabı ve açıklamayı göster

Cevap: $10$

Cevap

The value of

x

is

10

.

For five numbers arranged in ascending order with

x > 8

, the middle position (3rd value) is

8

, so the median is

8

. Summing all five values gives

3 + 7 + 8 + 12 + x = 30 + x

. Equating the mean

\frac{30 + x}{5}

to the median

8

yields

30 + x = 40

, which gives

x = 10

.

Adım Adım Çözüm

1

Determine the median of the ordered data set.

Since there are

5

numbers arranged in ascending order (

3, 7, 8, x, 12

) with

x > 8

, the median is the middle (3rd) number, which is

8

.

For an odd count of data points

n = 5

, the median position is

\frac{5+1}{2} = 3

.

2

Formulate the expression for the arithmetic mean.

\text{Mean} = \frac{3 + 7 + 8 + 12 + x}{5} = \frac{30 + x}{5}

.

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

3

Equate the mean to the median and solve for

x

.

\begin{align*} \frac{30 + x}{5} &= 8 \\ 30 + x &= 40 \\ x &= 10 \end{align*}

The question specifies that the mean of the data set is equal to its median.

Anahtar Kavram

Measures of Central Tendency for Ungrouped Data

Tahmini Süre:1m 30s

Soru 89Soru →

The frequency distribution table below shows the recorded speeds (in $\text{km/h}$ ) of a sample of commercial buses passing through a highway toll checkpoint:

Speed ( $\text{km/h}$ )	Frequency
$40 - 49$	$6$
$50 - 59$	$10$
$60 - 69$	$f$
$70 - 79$	$12$
$80 - 89$	$8$

If the mean speed of the buses is $66.5\text{ km/h}$ , find the value of the missing frequency $f$ .

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The value of the missing frequency

f

is

6

.

Each class interval's midpoint is calculated by averaging its lower and upper limits. The total frequency is

\sum f = 36 + f

and the total sum of products is

\sum fx = 2382 + 64.5f

. Applying the grouped mean formula

\bar{x} = \frac{\sum fx}{\sum f} = 66.5

gives

2394 + 66.5f = 2382 + 64.5f

, which yields

2f = 12

, so

f = 6

.

Adım Adım Çözüm

1

Find the class midpoints (

x

) for each grouped interval.

Midpoints are

44.5, 54.5, 64.5, 74.5,

and

84.5

.

Midpoints represent the central values of each interval for grouped mean calculations.

2

Calculate expressions for

\sum f

and

\sum fx

.

\sum f = 36 + f

and

\sum fx = 2382 + 64.5f$.

Summing the frequencies and the products of midpoints and frequencies gives the components required for the mean equation.

3

Substitute known values into the mean formula and solve for

f

.

66.5(36 + f) = 2382 + 64.5f \implies 2f = 12 \implies f = 6$.

Equating the formula expression to the given mean of

66.5\text{ km/h}

allows solving for the unknown frequency.

Anahtar Kavram

Grouped Mean with Missing Frequency

Soru 90Soru →

The frequency table below shows the distribution of masses (in kg) of $20$ packages delivered to a warehouse:

Mass (kg)	Frequency ( $f$ )
10 – 14	3
15 – 19	5
20 – 24	8
25 – 29	4

What is the mean mass of the packages?

18.25\text{ kg}

19.50\text{ kg}

20.25\text{ kg}

22.25\text{ kg}

Cevabı ve açıklamayı göster

Cevap: $20.25\text{ kg}$

Cevap

The mean mass of the packages is

20.25\text{ kg}

.

The mean of grouped frequency distribution is found using the formula

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

, where

x

represents the midpoint of each class interval. Calculating the midpoints yields

12

,

17

,

22

, and

27

. Multiplying each midpoint by its frequency gives products of

36

,

85

,

176

, and

108

, totaling

405

. Dividing

405

by the total frequency of

20

produces

20.25\text{ kg}

.

Adım Adım Çözüm

1

Find the class midpoint (

x

) for each class interval.

Midpoints are: 10–14

\rightarrow x = 12

; 15–19

\rightarrow x = 17

; 20–24

\rightarrow x = 22

; 25–29

\rightarrow x = 27

.

The mean of grouped data requires using the midpoint of each interval to represent its data values.

2

Multiply each class midpoint (

x

) by its frequency (

f

) to find

fx

.

3 \times 12 = 36

;

5 \times 17 = 85

;

8 \times 22 = 176

;

4 \times 27 = 108

.

This calculates the total estimated mass for each class interval.

3

Sum the frequencies (

\sum f

) and the products (

\sum fx

).

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

and

\sum fx = 36 + 85 + 176 + 108 = 405

.

These sums give the total number of packages and the total estimated mass.

4

Calculate the mean using the formula

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

.

\bar{x} = \frac{405}{20} = 20.25\text{ kg}

.

Dividing the total mass by the total number of items yields the grouped mean.

Anahtar Kavram

Mean of Grouped Data

Tahmini Süre:1m 0s

Soru 91Soru →

The table below shows the distribution of ages (in years) of trees in a forest reserve, presented alongside their frequency densities for a histogram representation:

Age Interval (years)	Frequency Density
$0 \le x < 10$	$3.0$
$10 \le x < 25$	$2.0$
$25 \le x < 45$	$2.25$
$45 \le x < 80$	$2.0$

If the dataset is instead displayed using a pie chart, what is the sector angle representing the age interval $25 \le x < 45$ ?

90^\circ

70^\circ

60^\circ

80^\circ

Cevabı ve açıklamayı göster

Cevap: $90^\circ$

Cevap

The sector angle representing the age interval

25 \le x < 45

is

90^\circ

.

The frequency of each interval is found by multiplying its frequency density by its class width (

3.0 \times 10 = 30

,

2.0 \times 15 = 30

,

2.25 \times 20 = 45

, and

2.0 \times 35 = 75

). The total frequency is

180

. The sector angle for

25 \le x < 45

(frequency

45

) is

\frac{45}{180} \times 360^\circ = 90^\circ

.

Adım Adım Çözüm

1

Calculate the class width for each interval.

Widths are:

10 - 0 = 10

,

25 - 10 = 15

,

45 - 25 = 20

, and

80 - 45 = 35

.

For histograms with unequal class widths, frequency density is defined as frequency divided by class width.

2

Calculate the frequency (

f

) for each interval using

f = \text{Frequency Density} \times \text{Class Width}

.

Interval

0 \le x < 10

:

f_1 = 3.0 \times 10 = 30

.
Interval

10 \le x < 25

:

f_2 = 2.0 \times 15 = 30

.
Interval

25 \le x < 45

:

f_3 = 2.25 \times 20 = 45

.
Interval

45 \le x < 80

:

f_4 = 2.0 \times 35 = 75

.

The frequency of a class in a histogram corresponds to the area of its bar.

3

Compute the total frequency (

N

).

N = 30 + 30 + 45 + 75 = 180

.

The total frequency represents the entire dataset needed for pie chart sector calculations.

4

Calculate the sector angle (

\,\theta\,

) for the interval

25 \le x < 45

.

\theta = \frac{f_3}{N} \times 360^\circ = \frac{45}{180} \times 360^\circ = 90^\circ

.

The sector angle in a pie chart is proportional to the relative frequency of the class out of

360^\circ

.

Anahtar Kavram

Conversion between Histogram Frequency Density and Pie Chart Sector Angles

Tahmini Süre:2m 0s

Soru 92Soru →

The table below presents the cumulative frequency distribution of examination marks for $60$ candidates:

Mark Boundary	Cumulative Frequency
$< 20.5$	$5$
$< 40.5$	$18$
$< 60.5$	$42$
$< 80.5$	$54$
$< 100.5$	$60$

How many candidates scored between $40.5$ and $80.5$ marks?

Cevabı ve açıklamayı göster

Cevap: 36

Cevap

36 candidates

To find the number of candidates with scores between

40.5

and

80.5

, subtract the cumulative frequency of scores below

40.5

(

18

) from the cumulative frequency of scores below

80.5

(

54

). This gives

54 - 18 = 36

.

Adım Adım Çözüm

1

Find the cumulative frequency up to the upper boundary of

80.5

Cumulative frequency (

F_{\text{upper}}

) =

54

This represents the total number of candidates scoring below

80.5

marks.

2

Find the cumulative frequency up to the lower boundary of

40.5

Cumulative frequency (

F_{\text{lower}}

) =

18

This represents the total number of candidates scoring below

40.5

marks.

3

Calculate the number of candidates within the interval

(40.5, 80.5)

by finding the difference

54 - 18 = 36

Subtracting the cumulative frequency at

40.5

from that at

80.5

isolates the count of candidates within this specific mark range.

Anahtar Kavram

Finding class frequency from cumulative frequency boundaries

Soru 93Soru →

A panel of $5$ members is to be selected from $10$ eligible candidates. If $2$ specific candidates refuse to serve on the panel together, in how many different ways can the panel be formed?

Cevabı ve açıklamayı göster

Cevap: 196

Cevap

The panel can be formed in 196 different ways.

To find the number of valid panels, use complementary counting. First, compute the total number of ways to pick any 5 candidates from 10 without restrictions:

\binom{10}{5} = 252

. Next, find the number of invalid panels that contain both restricted candidates; since 2 candidates are already placed, pick the remaining 3 members from the remaining 8 candidates:

\binom{8}{3} = 56

. Subtracting these invalid panels from the total gives

252 - 56 = 196

.

Adım Adım Çözüm

1

Calculate the total possible combinations without any restrictions.

\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252

The combinations formula

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

applies since the order of selection does not matter.

2

Determine the number of invalid combinations where both specific candidates are included.

\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

If both specific candidates are already placed on the panel, 3 more members must be chosen from the remaining 8 candidates.

3

Apply complementary counting to subtract invalid selections from total selections.

252 - 56 = 196

Subtracting the restricted combinations from the total combinations gives the number of valid panel configurations.

Anahtar Kavram

Combinations with mutual exclusion (Complementary Counting)

Soru 94Soru →

The table below shows the distribution of scores obtained by a group of candidates in an examination:

Score ( $x$ )	2	4	6	8	10
Frequency ( $f$ )	1	2	6	2	1

Calculate the variance of the scores.

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The variance of the scores is 4.

The mean score is

\bar{x} = 6

. Summing the weighted squared deviations gives

\sum f(x - 6)^2 = 48

. Dividing by the total frequency

N = 12

results in a variance of

\frac{48}{12} = 4

.

Adım Adım Çözüm

1

Find the total frequency and calculate the mean score.

Total frequency

N = 12

and mean

\bar{x} = 6

.

The mean is given by

\bar{x} = \frac{\sum f x}{\sum f} = \frac{2 + 8 + 36 + 16 + 10}{12} = \frac{72}{12} = 6

.

2

Determine the sum of squared deviations multiplied by their frequencies.

\sum f(x - \bar{x})^2 = 48.

Evaluating each term:

1(2-6)^2 = 16

,

2(4-6)^2 = 8

,

6(6-6)^2 = 0

,

2(8-6)^2 = 8

, and

1(10-6)^2 = 16

. Summing these yields

16 + 8 + 0 + 8 + 16 = 48

.

3

Compute the variance by dividing the total squared deviations by the total frequency.

Variance = 4.

Variance is

\sigma^2 = \frac{\sum f(x - \bar{x})^2}{\sum f} = \frac{48}{12} = 4

.

Anahtar Kavram

Variance of a Frequency Distribution

Soru 95Soru →

When constructing a standard cumulative frequency curve (ogive) for a grouped frequency distribution, against which statistical values on the horizontal axis are the cumulative frequencies plotted?

Upper class boundaries

Lower class boundaries

Class midpoints (class marks)

Raw class frequencies

Cevabı ve açıklamayı göster

Cevap: Upper class boundaries

Cevap

Upper class boundaries

In statistics, an ogive is constructed by plotting cumulative frequency on the vertical axis against the upper class boundaries of each interval on the horizontal axis.

Adım Adım Çözüm

1

Identify the definition of a cumulative frequency curve (ogive).

An ogive displays running totals of data values up to a given upper limit.

By definition, the cumulative frequency at any point represents all observations up to and including the upper boundary of that class.

2

Determine the correct variable for the horizontal axis.

The horizontal axis must represent the upper class boundaries (

x

-axis), while the vertical axis represents cumulative frequency (

y

-axis).

This ensures the curve accurately reflects cumulative proportions at the end of each class interval.

Anahtar Kavram

Plotting Cumulative Frequency Curves (Ogives)

Tahmini Süre:45s

Soru 96Soru →

The frequency distribution of scores $x$ obtained by a group of candidates in an aptitude test is given in the table below:

Score ( $x$ )	2	4	6	8	10
Frequency ( $f$ )	3	$k$	7	4	2

If the mean score of the distribution is $5.8$ , find the value of $k$ .

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The value of

k

is

4

.

The mean score of an ungrouped frequency distribution is given by

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

. Evaluating the sum of frequencies gives

\sum f = 16 + k

and the weighted sum of scores gives

\sum fx = 100 + 4k

. Substituting the given mean of

5.8

, we get

\frac{100 + 4k}{16 + k} = 5.8

. Cross-multiplying yields

100 + 4k = 92.8 + 5.8k

, leading to

1.8k = 7.2

, so

k = 4

.

Adım Adım Çözüm

1

Express total frequency

\sum f

in terms of

k

\sum f = 16 + k

The total number of observations is the sum of all frequencies in the distribution.

2

Calculate the sum of weighted scores

\sum fx

in terms of

k

\sum fx = 100 + 4k

Each score value must be multiplied by its corresponding frequency and summed together.

3

Set up the mean equation using the formula

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

5.8 = \frac{100 + 4k}{16 + k}

The mean of ungrouped data presented in a frequency table is total sum divided by total frequency.

4

Solve the linear equation for

k

k = 4

Cross-multiplying gives

92.8 + 5.8k = 100 + 4k

, which simplifies to

1.8k = 7.2

, yielding

k = 4

.

Anahtar Kavram

Mean of Ungrouped Data from a Frequency Table

Soru 97Soru →

A project team of $5$ members is to be formed from $6$ senior engineers and $4$ junior engineers. If the team must contain strictly more senior engineers than junior engineers, in how many different ways can the team be selected?

180

252

186

3600

Cevabı ve açıklamayı göster

Cevap: $186$

Cevap

186

To form a

5

-member team where senior engineers outnumber junior engineers, we consider three mutually exclusive cases: (5 senior, 0 junior), (4 senior, 1 junior), and (3 senior, 2 junior). Evaluating

^6C_5 \times ^4C_0 = 6

,

^6C_4 \times ^4C_1 = 60

, and

^6C_3 \times ^4C_2 = 120

, and adding them together yields

6 + 60 + 120 = 186

total ways.

Adım Adım Çözüm

1

Identify the conditions and valid cases for team composition

The team must have

5

members. The number of senior engineers must be strictly greater than junior engineers. The valid combinations of (senior, junior) engineers are:

(5, 0)

,

(4, 1)

, and

(3, 2)

.

Any other breakdown (such as 2 senior and 3 junior) would result in senior engineers not outnumbering junior engineers.

2

Calculate the combinations for Case 1 (

5

senior,

0

junior)

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

Choosing

5

out of

6

senior engineers and

0

out of

4

junior engineers.

3

Calculate the combinations for Case 2 (

4

senior,

1

junior)

^6C_4 \times ^4C_1 = 15 \times 4 = 60

Choosing

4

out of

6

senior engineers and

1

out of

4

junior engineers.

4

Calculate the combinations for Case 3 (

3

senior,

2

junior)

^6C_3 \times ^4C_2 = 20 \times 6 = 120

Choosing

3

out of

6

senior engineers and

2

out of

4

junior engineers.

5

Sum the combinations across all mutually exclusive cases

Total ways =

6 + 60 + 120 = 186

By the addition principle of counting, the total number of ways is the sum of ways from each disjoint case.

Anahtar Kavram

Conditional combinations using the formula

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

and the addition principle for disjoint cases.

Soru 98Soru →

The table below details the cumulative frequency distribution of the masses, in $\text{kg}$ , of $80$ agricultural packages recorded during an export inspection:

Mass Boundary ( $\text{kg}$ )	Cumulative Frequency
$\leq 19.5$	$8$
$\leq 29.5$	$24$
$\leq 39.5$	$52$
$\leq 49.5$	$72$
$\leq 59.5$	$80$

Using linear interpolation on the cumulative frequency data, determine the interquartile range of the package masses in $\text{kg}$ .

Cevabı ve açıklamayı göster

Cevap: 16.5

Cevap

The interquartile range of the package masses is

16.5\text{ kg}

.

The interquartile range measures the spread of the middle

50\%

of the data and is computed as

Q_3 - Q_1 = 43.5 - 27.0 = 16.5\text{ kg}

.

Adım Adım Çözüm

1

Find the lower quartile (

Q_1

)

Q_1 = 27.0\text{ kg}

The rank for

Q_1

is

\frac{1}{4} \times 80 = 20

, which lies in the

19.5 - 29.5

boundary interval with cumulative frequency increasing from

8

to

24

.

2

Find the upper quartile (

Q_3

)

Q_3 = 43.5\text{ kg}

The rank for

Q_3

is

\frac{3}{4} \times 80 = 60

, which lies in the

39.5 - 49.5

boundary interval with cumulative frequency increasing from

52

to

72

.

3

Calculate the Interquartile Range

\text{IQR} = 16.5\text{ kg}

Interquartile Range is the difference between the upper quartile (

Q_3

) and lower quartile (

Q_1

).

Anahtar Kavram

Interquartile Range from Cumulative Frequency Distribution

Soru 99Soru →

Two events $A$ and $B$ are defined in a sample space such that $P(A) = \frac{2}{3}$ , $P(B) = \frac{1}{4}$ , and $P(A \cup B) = \frac{3}{4}$ . If $B'$ denotes the complement of event $B$ , what is the value of $P(A \cap B')$ ?

\frac{1}{2}

\frac{1}{6}

\frac{2}{3}

\frac{1}{4}

Cevabı ve açıklamayı göster

Cevap: $\frac{1}{2}$

Cevap

The probability

P(A \cap B')

is

\frac{1}{2}

.

Using the addition law of probability,

P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{2}{3} + \frac{1}{4} - \frac{3}{4} = \frac{1}{6}

. Since

P(A) = P(A \cap B) + P(A \cap B')

, we find

P(A \cap B') = \frac{2}{3} - \frac{1}{6} = \frac{1}{2}

. Alternatively, since

P(A \cap B) = \frac{1}{6} = P(A)P(B)

, events

A

and

B

are independent, so

P(A \cap B') = P(A)P(B') = \frac{2}{3} \times \left(1 - \frac{1}{4}\right) = \frac{2}{3} \times \frac{3}{4} = \frac{1}{2}

.

Adım Adım Çözüm

1

Apply the addition law of probability to calculate

P(A \cap B)

.

P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{2}{3} + \frac{1}{4} - \frac{3}{4} = \frac{11}{12} - \frac{9}{12} = \frac{2}{12} = \frac{1}{6}

.

The general addition law relates the probabilities of the union and intersection of two compound events.

2

Calculate the probability that event

A

occurs while event

B

does not occur,

P(A \cap B')

.

P(A \cap B') = P(A) - P(A \cap B) = \frac{2}{3} - \frac{1}{6} = \frac{4}{6} - \frac{1}{6} = \frac{3}{6} = \frac{1}{2}

.

Event

A

can be decomposed into two mutually exclusive parts:

A \cap B

and

A \cap B'

.

Anahtar Kavram

Addition law of probability and complement of compound events

Tahmini Süre:2m 0s

Soru 100Soru →

A machine produces electronic components. Based on long-term specifications, the theoretical probability of producing a defective component is $0.04$ . In a quality control inspection, a random sample of $400$ components produced by the machine is tested, and $22$ are found to be defective. What is the absolute difference between the experimental probability and the theoretical probability of selecting a defective component from this sample?

Cevabı ve açıklamayı göster

Cevap: 0.015

Cevap

The absolute difference between the experimental probability and theoretical probability is 0.015.

The experimental probability is calculated from the trial outcomes as 22 divided by 400, which equals 0.055. Taking the absolute difference between this value and the theoretical probability of 0.04 gives |0.055 - 0.04| = 0.015.

Adım Adım Çözüm

1

Determine the experimental probability from the sample results

P(Experimental) = \frac{22}{400} = 0.055

Experimental probability is calculated as the ratio of observed successful outcomes (defective items found) to total trials (sample size).

2

Identify the given theoretical probability

P(Theoretical) = 0.04

Theoretical probability represents the expected likelihood under model specifications prior to sampling.

3

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

|0.055 - 0.04| = 0.015

The absolute difference measures how much the observed empirical frequency deviates from the predicted model probability.

Anahtar Kavram

Experimental vs Theoretical Probability

Statistics and Probability

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