Cumulative Frequency and Ogive Alıştırma Soruları — JAMB UTME

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 2Soru →

The table below shows the cumulative frequency distribution of the masses (in grams) of $100$ cocoa beans sampled from an agricultural yield:

Mass Class Interval (g)	Cumulative Frequency
$10 - 19$	$10$
$20 - 29$	$30$
$30 - 39$	$65$
$40 - 49$	$90$
$50 - 59$	$100$

Using linear interpolation from the cumulative frequency distribution, calculate the 75th percentile ( $Q_3$ ) mass of the cocoa beans in grams.

Cevabı ve açıklamayı göster

Cevap: 43.5

Cevap

The 75th percentile (

Q_3

) mass of the cocoa beans is

43.5\text{ g}

.

To find the 75th percentile (

Q_3

) from cumulative frequency data, calculate the rank

\frac{75}{100} \times 100 = 75

. This falls into the

40 - 49

class interval (boundaries

39.5 - 49.5

). Applying

Q_3 = L + \left(\frac{75 - F}{f}\right)c

yields

39.5 + \left(\frac{75 - 65}{25}\right) \times 10 = 43.5\text{ g}

.

Adım Adım Çözüm

1

Calculate the percentile position rank

Rank position is

75

The 75th percentile corresponds to

75\%

of the total sample size

N = 100

, giving

\frac{75}{100} \times 100 = 75

.

2

Locate the 75th percentile class interval and its boundaries

Class interval is

40 - 49

, with lower boundary

L = 39.5

and upper boundary

49.5

Cumulative frequency before

40 - 49

is

65

, and up to

40 - 49

is

90

. Since

65 < 75 \leq 90

, the 75th item falls in this interval.

3

Identify class parameters for interpolation

L = 39.5

,

F = 65

,

f = 25

,

c = 10

Lower boundary

L = 39.5

, previous cumulative frequency

F = 65

, class frequency

f = 90 - 65 = 25

, class width

c = 49.5 - 39.5 = 10

.

4

Compute

Q_3

using the linear interpolation formula

Q_3 = 43.5\text{ g}

Q_3 = 39.5 + \left(\frac{75 - 65}{25}\right) \times 10 = 39.5 + 4 = 43.5

.

Anahtar Kavram

Linear Interpolation of Percentiles from Cumulative Frequency Data

Soru 3Soru →

The frequency distribution table below shows the daily rainfall (in mm) recorded across $80$ weather monitoring stations during a storm:

Daily Rainfall (mm)	Frequency ( $f$ )
$10 - 19$	$8$
$20 - 29$	$14$
$30 - 39$	$26$
$40 - 49$	$20$
$50 - 59$	$12$

Using linear interpolation from the cumulative frequency distribution (ogive), what is the 75th percentile ( $P_{75}$ ) of the daily rainfall in mm?

Cevabı ve açıklamayı göster

Cevap: 45.5

Cevap

The 75th percentile of the daily rainfall distribution is

45.5\text{ mm}

.

To find the 75th percentile (

P_{75}

), determine the

60^{\text{th}}

cumulative frequency position (

0.75 \times 80 = 60

). The value lies within the

40 - 49

class interval. Applying the lower class boundary

L = 39.5

, preceding cumulative frequency

cf_b = 48

, frequency

f = 20

, and class width

c = 10

, linear interpolation yields

P_{75} = 39.5 + \frac{60 - 48}{20} \times 10 = 45.5\text{ mm}

.

Adım Adım Çözüm

1

Calculate cumulative frequencies across all class intervals.

Cumulative frequencies are

8

for

10-19

,

22

for

20-29

,

48

for

30-39

,

68

for

40-49

, and

80

for

50-59

. Total frequency

N = 80

.

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

2

Determine the position corresponding to the 75th percentile.

Position

= 0.75 \times 80 = 60^{\text{th}}

cumulative frequency item.

The 75th percentile represents

75\%

of the total sample size.

3

Identify the target class interval parameters containing the 60th observation.

The interval

40 - 49

contains cumulative frequencies from

49

to

68

. Parameters:

L = 39.5

,

c = 10

,

f = 20

,

cf_b = 48

.

Linear interpolation requires the exact boundaries and frequencies of the container class.

4

Apply the percentile interpolation formula

P_{75} = L + \left(\frac{60 - cf_b}{f}\right) \times c

.

P_{75} = 39.5 + \left(\frac{60 - 48}{20}\right) \times 10 = 39.5 + 6 = 45.5\text{ mm}

.

Calculates the exact rainfall value corresponding to the 75th percentile.

Anahtar Kavram

Linear interpolation for percentiles using cumulative frequency distribution (ogive)

Tahmini Süre:1m 30s

Soru 4Soru →

The cumulative frequency distribution table below shows the completion times (in minutes) for $120$ software engineers during a technical skill test:

Time Interval (min)	Frequency ( $f$ )	Cumulative Frequency ( $cf$ )
$10 - 19$	$12$	$12$
$20 - 29$	$28$	$40$
$30 - 39$	$40$	$80$
$40 - 49$	$24$	$104$
$50 - 59$	$16$	$120$

Using linear interpolation from the cumulative frequency data, what is the $65^{\text{th}}$ percentile completion time?

39.0\text{ minutes}

29.0\text{ minutes}

39.5\text{ minutes}

34.5\text{ minutes}

Cevabı ve açıklamayı göster

Cevap: $39.0\text{ minutes}$

Cevap

39.0\text{ minutes}

To find the

65^{\text{th}}

percentile (

P_{65}

), we calculate

\frac{65}{100} \times 120 = 78

. The cumulative frequency table shows that rank

78

falls within the class interval

30 - 39

. The true lower boundary for this class is

L = 29.5

, the cumulative frequency of the preceding class is

F = 40

, the frequency of the percentile class is

f = 40

, and the class width is

c = 10

. Substituting into

P_{65} = L + \left(\frac{78 - F}{f}\right)c

yields

29.5 + \left(\frac{38}{40}\right) \times 10 = 39.0\text{ minutes}

.

Adım Adım Çözüm

1

Determine the rank of the

65^{\text{th}}

percentile (

P_{65}

).

Rank

= \frac{65}{100} \times 120 = 78

.

The percentile rank identifies the position of the data point within the total frequency

N = 120

.

2

Identify the percentile class and its boundaries.

Percentile class is

30 - 39

, with lower boundary

L = 29.5

, upper boundary

= 39.5

, and class width

c = 10

.

The cumulative frequency increases from

40

to

80

across this interval, containing rank

78

.

3

Apply the linear interpolation formula for percentiles on grouped data.

P_{65} = L + \left(\frac{\frac{65N}{100} - F}{f}\right) \times c = 29.5 + \left(\frac{78 - 40}{40}\right) \times 10 = 29.5 + 9.5 = 39.0\text{ minutes}

.

Where

F = 40

is the cumulative frequency prior to the class and

f = 40

is the frequency of the class.

Anahtar Kavram

Linear Interpolation of Percentiles from Cumulative Frequency Distributions

Soru 5Soru →

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 6Soru →

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 7Soru →

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 8Soru →

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 9Soru →

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 10Soru →

The cumulative frequency distribution of the operational lifespans (in hours) for a batch of $100$ precision LED modules tested in a laboratory is summarized below:

Lifespan Interval (hours)	Class Boundaries	Cumulative Frequency
$100 - 119$	$99.5 - 119.5$	$10$
$120 - 139$	$119.5 - 139.5$	$25$
$140 - 159$	$139.5 - 159.5$	$60$
$160 - 179$	$159.5 - 179.5$	$85$
$180 - 199$	$179.5 - 199.5$	$100$

Using linear interpolation for cumulative frequency distributions, calculate the 75th percentile ( $P_{75}$ ) of the lifespan of these modules in hours.

Cevabı ve açıklamayı göster

Cevap: 171.5

Cevap

The 75th percentile of operational lifespan is 171.5 hours.

The 75th percentile rank position is

0.75 \times 100 = 75

. The percentile falls within the class boundary

159.5 - 179.5

. Substituting lower class boundary

L = 159.5

, preceding cumulative frequency

c.f. = 60

, class frequency

f = 25

, and class width

c = 20

into

P_{75} = L + \left(\frac{75 - c.f.}{f}\right) \times c

gives

159.5 + \left(\frac{15}{25}\right) \times 20 = 171.5

hours.

Adım Adım Çözüm

1

Determine the rank position of the 75th percentile.

Rank position = 75th value out of 100.

The 75th percentile corresponds to 75% of the total frequency N = 100.

2

Identify the percentile class interval and extract relevant parameters.

Class interval is 159.5 - 179.5, with L = 159.5, c.f. = 60, f = 25, and c = 20.

The cumulative frequency increases from 60 to 85 across the boundary 159.5 to 179.5, which contains the 75th value.

3

Compute the percentile value using ogive linear interpolation.

P_75 = 159.5 + [(75 - 60) / 25] * 20 = 171.5 hours.

Applying the cumulative frequency interpolation formula yields the exact value.

Anahtar Kavram

Calculating Percentiles from Cumulative Frequency / Ogives

Soru 11Soru →

The table below presents the cumulative frequency distribution of delay times (in minutes) for $50$ regional flights recorded at an airport:

Delay Time (minutes)	Cumulative Frequency
$10 - 19$	$8$
$20 - 29$	$20$
$30 - 39$	$40$
$40 - 49$	$50$

What is the $70\text{th}$ percentile of the delay times?

37.0

minutes

37.5

minutes

32.0

minutes

35.0

minutes

Cevabı ve açıklamayı göster

Cevap: $37.0$ minutes

Cevap

The

70\text{th}

percentile of the delay times is

37.0

minutes.

The

70\text{th}

percentile position is found by taking

70\%

of

50

, which yields

35

. Looking at the cumulative frequency table, the 35th item falls in the

30 - 39

interval (class boundaries

29.5 - 39.5

). Applying the percentile interpolation formula:

P_{70} = 29.5 + \left(\frac{35 - 20}{20}\right) \times 10 = 29.5 + 7.5 = 37.0\text{ minutes}

.

Adım Adım Çözüm

1

Determine the position of the 70th percentile rank.

Position rank

= \frac{70}{100} \times 50 = 35\text{th value}

.

The 70th percentile corresponds to 70% of the total frequency

N = 50

.

2

Identify the 70th percentile class interval and its parameters.

Interval:

30 - 39

; Lower class boundary

L = 29.5

, cumulative frequency preceding class

F = 20

, frequency of percentile class

f = 20

, class width

c = 10

.

The cumulative frequency reaches 20 at

29.5

and 40 at

39.5

, so the 35th value lies in the

30 - 39

class interval.

3

Apply the percentile interpolation formula

P_{70} = L + \left(\frac{\frac{70N}{100} - F}{f}\right) \times c

.

P_{70} = 29.5 + \left(\frac{35 - 20}{20}\right) \times 10 = 29.5 + \left(\frac{15}{20}\right) \times 10 = 29.5 + 7.5 = 37.0\text{ minutes}

.

Linear interpolation distributes the cumulative frequencies evenly across the class interval.

Anahtar Kavram

Calculating percentiles from grouped cumulative frequency data using class boundaries and linear interpolation.

Soru 12Soru →

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

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

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

Cevabı ve açıklamayı göster

Cevap: 137.5

Cevap

137.5 kWh

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

0.60 \times 50 = 30

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

100

and

150

. Using the lower boundary of

100

, previous cumulative frequency of

15

, interval frequency of

20

, and interval width of

50

, linear interpolation gives

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

kWh.

Adım Adım Çözüm

1

Determine the rank for the 60th percentile

Rank position = 30th household

The 60th percentile rank corresponds to

60\%

of the total sample size (

N = 50

), which is

0.60 \times 50 = 30

.

2

Identify the parameters of the percentile class interval

Lower boundary

L = 100

,

CF_{\text{prev}} = 15

, class frequency

f = 20

, width

w = 50

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

(100, 150]

, so the 30th value lies within this interval.

3

Calculate the value using linear interpolation

137.5 kWh

Substitute values into

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

kWh.

Anahtar Kavram

Linear interpolation for percentiles from a cumulative frequency distribution

Soru 13Soru →

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

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

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

Cevabı ve açıklamayı göster

Cevap: 35

Cevap

The median mass of cocoa beans harvested by the farmers is

35.0\text{ kg}

.

The total number of farmers is

N = 40

. The median corresponds to the

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

position. From the cumulative frequency table, the

20^{\text{th}}

item lies within the

30 - 40\text{ kg}

class interval. Using the grouped median formula

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

, where

L = 30

,

c.f. = 12

,

f = 28 - 12 = 16

, and

w = 10

, we obtain

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

.

Adım Adım Çözüm

1

Find the median position in the cumulative frequency distribution

Median position

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

item

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

N = 40

.

2

Locate the median class interval and extract its statistical parameters

Median class interval is

30 - 40\text{ kg}

, with lower limit

L = 30\text{ kg}

, preceding cumulative frequency

c.f. = 12

, class frequency

f = 28 - 12 = 16

, and width

w = 10\text{ kg}

The cumulative frequency just below

20

is

12

(at upper boundary

30

), and at upper boundary

40

it rises to

28

.

3

Substitute values into the grouped data median formula

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

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

Anahtar Kavram

Calculation of Median from Cumulative Frequency Data

Soru 14Soru →

The table below shows the distribution of examination scores of $50$ candidates in a selection test:

Score Class	Frequency ( $f$ )
$10 - 19$	$5$
$20 - 29$	$9$
$30 - 39$	$16$
$40 - 49$	$14$
$50 - 59$	$6$

Using the cumulative frequency distribution (ogive), what is the estimated $70^{\text{th}}$ percentile score?

33.1

43.1

43.6

38.1

Cevabı ve açıklamayı göster

Cevap: $43.1$

Cevap

The estimated

70^{\text{th}}

percentile score is

43.1

.

The

70^{\text{th}}

percentile corresponds to the score at the

35^{\text{th}}

candidate (

70\%

of

50

). This falls within the

40 - 49

score class, which has a lower boundary of

39.5

, a frequency of

14

, and a class width of

10

. Interpolating linearly gives

39.5 + \left(\frac{35 - 30}{14}\right) \times 10 = 43.07 \approx 43.1

.

Adım Adım Çözüm

1

Construct the cumulative frequency table to find class boundaries and cumulative frequencies.

Cumulative frequencies:

10-19

(

CF = 5

),

20-29

(

CF = 14

),

30-39

(

CF = 30

),

40-49

(

CF = 44

),

50-59

(

CF = 50

).

Cumulative frequencies are required to locate the position of the desired percentile.

2

Determine the rank position of the

70^{\text{th}}

percentile (

P_{70}

).

Rank position =

\frac{70}{100} \times 50 = 35^{\text{th}}

position.

The

70^{\text{th}}

percentile corresponds to the score below which

70\%

of the total candidates fall.

3

Identify the class interval containing the

35^{\text{th}}

cumulative frequency and state its parameters.

The target class is

40 - 49

. Parameters: Lower class boundary

L = 39.5

, preceding cumulative frequency

CF_{\text{prev}} = 30

, class frequency

f = 14

, class width

c = 10

.

Since

30 < 35 \le 44

, the

35^{\text{th}}

entry falls within the

40 - 49

class.

4

Apply the linear interpolation formula for percentiles from an ogive.

P_{70} = L + \left(\frac{\frac{70N}{100} - CF_{\text{prev}}}{f}\right) \times c = 39.5 + \left(\frac{35 - 30}{14}\right) \times 10 = 39.5 + \frac{50}{14} \approx 43.07 \approx 43.1

.

This calculates the exact score estimate on the cumulative frequency curve.

Anahtar Kavram

Estimating Percentiles from Grouped Data and Ogives

Cumulative Frequency and Ogive

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