Frequency Distributions, Tables, and Grouped Data Practice Questions — GRE General Test

Question 1Question →

The table below shows the frequency distribution of daily passenger counts (in hundreds) for a city bus route recorded over a 40-day period.

Daily Passengers (in hundreds)	Frequency (Number of Days)
$10 - 14$	$5$
$15 - 19$	$12$
$20 - 24$	$15$
$25 - 29$	$8$

If the mean of the grouped data is estimated by using the midpoint of each class interval, what is the estimated mean daily passenger count (in hundreds)?

Show answer & explanation

Answer: 20.25

Answer

The estimated mean daily passenger count is 20.25 hundred passengers.

To calculate the estimated mean of grouped data, find the midpoint of each interval, multiply each midpoint by its frequency, sum those products (

810

), and divide by the total number of observations (

40

). This gives

\frac{810}{40} = 20.25

.

Step-by-Step Solution

1

Calculate the midpoints for each of the four class intervals.

The midpoints are

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

,

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

,

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

, and

\frac{25+29}{2} = 27

.

To estimate the mean of grouped frequency data, each interval is represented by its center value (midpoint).

2

Multiply each class midpoint by its frequency and calculate the total sum of these products.

(12 \times 5) + (17 \times 12) + (22 \times 15) + (27 \times 8) = 60 + 204 + 330 + 216 = 810

.

Multiplying each midpoint by its frequency computes the total estimated value contributed by all observations in that class.

3

Divide the total estimated value by the total sample size (total frequency).

\frac{810}{40} = 20.25

.

The weighted average (grouped mean) is the sum of weighted midpoints divided by the total frequency.

Key Concept

Estimated Mean of Grouped Data

Question 2Question →

The frequency distribution table below summarizes the processing times, in minutes, for a sample of 100 health insurance claims processed by an administrative team.

Claim Processing Time (minutes)	Frequency
$10 \le t < 20$	15
$20 \le t < 30$	25
$30 \le t < 40$	35
$40 \le t < 50$	20
$50 \le t < 60$	5

Which of the following statements regarding this frequency distribution must be true? Select all that apply.

Select all that apply

The interval

30 \le t < 40

contains the median processing time for the dataset.

Exactly 60 percent of the claims required 30 minutes or more to process.

The lower bound estimate for the overall mean processing time is 31.0 minutes.

The number of claims in the interval

20 \le t < 30

represents 25 percent of all claims.

The ratio of the number of claims completed in under 30 minutes to the number of claims taking 40 minutes or more is 2 to 1.

Show answer & explanation

Answer: The interval $30 \le t < 40$ contains the median processing time for the dataset.; Exactly 60 percent of the claims required 30 minutes or more to process.; The number of claims in the interval $20 \le t < 30$ represents 25 percent of all claims.

Answer

The true statements are: the median processing time lies in the interval

30 \le t < 40

; exactly 60 percent of the claims required 30 minutes or more to process; and the number of claims in the interval

20 \le t < 30

represents 25 percent of all claims.

The statement regarding the median is correct because the 50th and 51st values lie within the cumulative frequency span of 41 to 75, corresponding to

30 \le t < 40

. The statement regarding claims requiring 30 minutes or more is correct because

35 + 20 + 5 = 60

out of 100 claims is 60 percent. The statement regarding the

20 \le t < 30

interval is correct because 25 out of 100 claims is exactly 25 percent.

Step-by-Step Solution

1

Calculate cumulative frequencies to locate the median class interval.

Cumulative frequencies are 15 (

10 \le t < 20

), 40 (

20 \le t < 30

), 75 (

30 \le t < 40

), 95 (

40 \le t < 50

), and 100 (

50 \le t < 60

). For 100 observations, the 50th and 51st values fall in the

30 \le t < 40

interval.

The median of an even number

N=100

of sorted values is the average of the 50th and 51st values.

2

Calculate the percentage of claims taking 30 minutes or more.

Sum the frequencies for intervals

\ge 30

:

35 + 20 + 5 = 60

. As a percentage of 100 total claims, this is

\frac{60}{100} \times 100\% = 60\%

.

To find the proportion meeting or exceeding a threshold, sum all class frequencies with lower limits at or above 30 minutes.

3

Evaluate the percentage representation of the

20 \le t < 30

interval.

Frequency is 25 out of 100 total claims, which equals

\frac{25}{100} = 25\%

.

The relative frequency of a grouped class is its frequency divided by the total frequency.

4

Determine the lower bound for the estimated mean processing time to evaluate incorrect claims.

Minimum sum =

15(10) + 25(20) + 35(30) + 20(40) + 5(50) = 2750

. Lower bound mean =

2750 / 100 = 27.5

minutes.

The absolute minimum mean occurs when every observation equals the lower boundary of its respective interval.

5

Calculate the ratio of claims under 30 minutes to claims 40 minutes or more.

Claims

< 30

min =

15 + 25 = 40

. Claims

\ge 40

min =

20 + 5 = 25

. Ratio =

40 : 25 = 8 : 5

.

Must include all relevant intervals (including

50 \le t < 60

) when computing the denominator for

\ge 40

minutes.

Key Concept

Grouped Frequency Distributions, Median Class Identification, and Relative Frequencies

Question 3Question →

The frequency distribution table below shows the daily electricity consumption, in kilowatt-hours ( $\text{kWh}$ ), recorded for a sample of $40$ small retail stores over a monitoring period.

Daily Electricity Consumption ( $\text{kWh}$ )	Number of Stores
$10 \le e < 30$	$6$
$30 \le e < 50$	$14$
$50 \le e < 70$	$12$
$70 \le e < 90$	$8$

Based on the table, what is the estimated mean daily electricity consumption, in $\text{kWh}$ , per store, using the midpoint of each interval to represent all data within that interval?

Show answer & explanation

Answer: 51

Answer

51

To estimate the mean from grouped frequency data, each interval is represented by its midpoint. The midpoints for the four intervals are

20

,

40

,

60

, and

80

. Multiplying each midpoint by its frequency gives

20(6) = 120

,

40(14) = 560

,

60(12) = 720

, and

80(8) = 640

. The sum of these products is

120 + 560 + 720 + 640 = 2040

. Dividing this sum by the total frequency of

40

stores gives an estimated mean of

51\text{ kWh}

.

Step-by-Step Solution

1

Determine the class midpoint for each interval of electricity consumption.

Midpoints:

\frac{10+30}{2} = 20

,

\frac{30+50}{2} = 40

,

\frac{50+70}{2} = 60

, and

\frac{70+90}{2} = 80

.

When working with grouped frequency data, each interval is represented by its midpoint value to estimate total sums.

2

Multiply each midpoint by the number of stores in that interval to find the estimated sum of consumption per group.

20 \times 6 = 120

,

40 \times 14 = 560

,

60 \times 12 = 720

,

80 \times 8 = 640

.

Weighting each midpoint by its interval frequency yields the total estimated consumption across all stores in that bin.

3

Sum the total estimated consumption across all intervals.

120 + 560 + 720 + 640 = 2040 \text{ kWh}

.

This calculation aggregate gives the total estimated electricity consumed by all 40 stores combined.

4

Divide the total estimated electricity consumption by the total number of stores.

\frac{2040}{40} = 51 \text{ kWh}

.

Dividing the aggregate total by the sample size gives the estimated arithmetic mean per store.

Key Concept

Calculating the estimated mean of grouped data using class midpoints and weighted frequencies.

Estimated Time:1m 30s

Question 4Question →

The table below shows the frequency distribution of the weights, in grams, for a sample of 80 manufactured components.

Weight (grams)	Frequency
100–109	12
110–119	24
120–129	28
130–139	16

What percentage of the components in the sample have a weight of at least 120 grams?

Show answer & explanation

Answer: 55

Answer

55%

To calculate the percentage of components with a weight of at least 120 grams, sum the frequencies of all weight intervals corresponding to 120 grams or more:

28 + 16 = 44

. Next, divide this count by the total number of components (

80

) and multiply by

100

:

\frac{44}{80} \times 100\% = 55\%

.

Step-by-Step Solution

1

Identify the number of components with weight at least 120 grams

44 components

The weight intervals '120–129' and '130–139' contain 28 and 16 components respectively, giving

28 + 16 = 44

.

2

Calculate the percentage relative to the total sample size of 80 components

55%

Dividing the target count (44) by the total sample size (80) and multiplying by 100 yields

\frac{44}{80} \times 100 = 55\%

.

Key Concept

Calculating class percentages from grouped frequency tables

Question 5Question →

The table below shows the frequency distribution of customer satisfaction ratings collected by a service center over a 50-day monitoring period.

Rating Interval	Frequency (Days)
1–5	8
6–10	12
11–15	20
16–20	10

Based on the table, what percentage of the days had a customer satisfaction rating of 11 or higher?

20%

40%

50%

60%

75%

Show answer & explanation

Answer: 60%

Answer

60%

To find the percentage of days with a customer satisfaction rating of 11 or higher, add the frequencies of the intervals '11–15' (20 days) and '16–20' (10 days) to get 30 days. The total number of days across all intervals is 8 + 12 + 20 + 10 = 50 days. The percentage is (30 / 50) × 100% = 60%.

Step-by-Step Solution

1

Identify the relevant rating intervals.

The intervals corresponding to a rating of 11 or higher are '11–15' and '16–20'.

Ratings of 11 or higher include all data points within these two top intervals.

2

Sum the frequencies for the relevant intervals.

Frequency = 20 + 10 = 30 days.

Combining the frequencies gives the total number of favorable outcomes.

3

Calculate the total number of days across all intervals.

Total days = 8 + 12 + 20 + 10 = 50 days.

The total frequency represents the denominator for the percentage calculation.

4

Compute the percentage.

(30 / 50) × 100% = 60%.

Dividing the target frequency by the total frequency and multiplying by 100 yields the required percentage.

Key Concept

Frequency Table Percentages and Grouped Data Cutoffs

Estimated Time:45s

Question 6Question →

The table below shows the frequency distribution of daily rainfall, in millimeters, recorded across 50 weather monitoring stations during a storm event.

Daily Rainfall (mm)	Number of Stations
0 to under 10	10
10 to under 20	15
20 to under 30	17
30 to under 40	5
40 to under 50	3

What percentage of the stations that recorded at least 10 mm of daily rainfall recorded less than 30 mm of daily rainfall?

42.5\%

64\%

75\%

80\%

84\%

Show answer & explanation

Answer: $80\%$

Answer

80\%

To find the desired percentage, first identify the total number of stations that recorded at least 10 mm of rainfall. This includes the intervals 10 to under 20, 20 to under 30, 30 to under 40, and 40 to under 50, giving

15 + 17 + 5 + 3 = 40

stations. Next, find how many of these 40 stations recorded less than 30 mm. These fall into the 10 to under 20 and 20 to under 30 intervals, which total

15 + 17 = 32

stations. Calculating the relative percentage gives

(32 / 40) \times 100\% = 80\%

.

Step-by-Step Solution

1

Determine the conditional base (denominator)

Sum the frequencies for all intervals representing at least 10 mm:

15 + 17 + 5 + 3 = 40

stations.

The question specifies 'stations that recorded at least 10 mm of daily rainfall', restricting our population base to this subset.

2

Determine the target subgroup frequency (numerator)

Sum the frequencies for intervals within the base that are also less than 30 mm (the 10 to under 20 and 20 to under 30 intervals):

15 + 17 = 32

stations.

We need to count the stations that meet both conditions: at least 10 mm and less than 30 mm.

3

Calculate the percentage

\frac{32}{40} \times 100\% = 0.80 \times 100\% = 80\%

.

Divide the target subgroup count by the conditional base count and multiply by 100.

Key Concept

Grouped frequency table interpretation with conditional percentages

Estimated Time:1m 30s

Question 7Question →

The table below shows the frequency distribution of order fulfillment times, in minutes, for a sample of $50$ online orders processed at a warehouse facility.

Fulfillment Time $t$ (minutes)	Number of Orders
$10 \le t < 20$	$10$
$20 \le t < 30$	$15$
$30 \le t < 40$	$15$
$40 \le t < 50$	$7$
$50 \le t < 60$	$3$

An order is categorized as "delayed" if its fulfillment time is at least $30$ minutes. What percentage of all orders that took at least $20$ minutes to fulfill were categorized as "delayed"?

30.0%

37.5%

50.0%

62.5%

75.0%

Show answer & explanation

Answer: 62.5%

Answer

62.5%

To find the required percentage, first determine the subset of orders that took at least

20

minutes to fulfill. Summing the frequencies of the last four intervals (

20 \le t < 30

,

30 \le t < 40

,

40 \le t < 50

, and

50 \le t < 60

) yields

15 + 15 + 7 + 3 = 40

orders. Next, find how many of these orders took at least

30

minutes to fulfill (the delayed orders), which corresponds to the last three intervals:

15 + 7 + 3 = 25

orders. Finally, calculate the percentage:

\frac{25}{40} \times 100\% = 62.5\%

.

Step-by-Step Solution

1

Identify the sub-population that defines the base (denominator) of the percentage.

The target base is all orders with fulfillment times of at least

20

minutes (

t \ge 20

). Summing the frequencies for intervals

20 \le t < 30

,

30 \le t < 40

,

40 \le t < 50

, and

50 \le t < 60

gives

15 + 15 + 7 + 3 = 40

orders.

The question specifically asks 'of all orders that took at least 20 minutes to fulfill', establishing a conditional base.

2

Determine the number of orders meeting the numerator condition.

The delayed orders are those taking at least

30

minutes (

t \ge 30

). Summing frequencies for

30 \le t < 40

,

40 \le t < 50

, and

50 \le t < 60

gives

15 + 7 + 3 = 25

orders.

Orders in all intervals starting from

30

minutes onwards qualify as delayed.

3

Compute the required percentage.

\left(\frac{25}{40}\right) \times 100\% = 0.625 \times 100\% = 62.5\%

.

Dividing the target count by the restricted base count gives the conditional proportion.

Key Concept

Grouped Frequency Distributions and Conditional Percentages

Question 8Question →

An urban forestry department measured the trunk diameters, in inches, of a sample of $60$ oak trees in a municipal park. The table below summarizes the frequency distribution of the measurements.

Trunk Diameter (inches)	Frequency
$10$ to $14$	$8$
$15$ to $19$	$15$
$20$ to $24$	$22$
$25$ to $29$	$11$
$30$ to $34$	$4$

Based on the table, what fraction of the trees with a trunk diameter of at least $15$ inches have a trunk diameter of less than $25$ inches?

\frac{37}{60}

\frac{37}{52}

\frac{45}{52}

\frac{11}{26}

\frac{37}{15}

Show answer & explanation

Answer: $\frac{37}{52}$

Answer

The fraction of trees with a trunk diameter of at least 15 inches that have a trunk diameter of less than 25 inches is 37/52.

To calculate the required fraction, first identify the subset of trees that have a trunk diameter of at least 15 inches. Summing the frequencies of the intervals 15–19, 20–24, 25–29, and 30–34 gives 15 + 22 + 11 + 4 = 52 trees. Next, identify how many of these 52 trees have a trunk diameter of less than 25 inches. These are the trees in the intervals 15–19 and 20–24, giving 15 + 22 = 37 trees. Therefore, the fraction of trees meeting the condition is 37/52.

Step-by-Step Solution

1

Determine the total number of trees satisfying the condition 'at least 15 inches'.

Sum the frequencies for intervals 15–19, 20–24, 25–29, and 30–34:

15 + 22 + 11 + 4 = 52

trees.

This establishes the restricted denominator (base group) specified by the conditional statement.

2

Determine the number of trees within this group that satisfy the condition 'less than 25 inches'.

Sum the frequencies for intervals 15–19 and 20–24:

15 + 22 = 37

trees.

These are the target intervals that are both at least 15 inches and strictly less than 25 inches.

3

Calculate the required fraction.

\frac{37}{52}

.

Divide the target count by the conditional total count to form the part-to-whole ratio.

Key Concept

Conditional relative frequencies and part-to-whole calculations from grouped data tables.

Estimated Time:1m 30s

Question 9Question →

The frequency distribution table below summarizes the monthly water consumption, $c$ (in cubic meters, $\text{m}^3$ ), recorded for a sample of $150$ municipal water accounts.

Monthly Water Consumption ( $\text{m}^3$ )	Number of Accounts
$0 \le c < 10$	$25$
$10 \le c < 20$	$45$
$20 \le c < 30$	$50$
$30 \le c < 40$	$20$
$40 \le c < 50$	$10$

If one water account is selected at random from among all accounts with a monthly water consumption of at least $10\text{ m}^3$ , what is the probability that the selected account has a monthly water consumption of less than $30\text{ m}^3$ ? (Give your answer as a decimal rounded to two decimal places.)

Show answer & explanation

Answer: 0.76

Answer

0.76

To calculate the required probability, first restrict the sample space to accounts with a monthly consumption of at least

10\text{ m}^3

. Summing the frequencies for the intervals

10 \le c < 20

,

20 \le c < 30

,

30 \le c < 40

, and

40 \le c < 50

gives

45 + 50 + 20 + 10 = 125

accounts. Among these

125

accounts, those with a consumption of less than

30\text{ m}^3

fall into the intervals

10 \le c < 20

and

20 \le c < 30

, giving a count of

45 + 50 = 95

accounts. Dividing the favorable outcomes by the total outcomes in the restricted sample space yields

\frac{95}{125} = 0.76

.

Step-by-Step Solution

1

Determine the total number of accounts meeting the condition of having consumption of at least

10\text{ m}^3

.

Total eligible accounts =

45 + 50 + 20 + 10 = 125

.

Accounts with consumption of at least

10\text{ m}^3

fall into the intervals

10 \le c < 20

,

20 \le c < 30

,

30 \le c < 40

, and

40 \le c < 50

.

2

Determine the number of accounts among the eligible set with consumption less than

30\text{ m}^3

.

Number of favorable accounts =

45 + 50 = 95

.

Within the eligible set, accounts with consumption less than

30\text{ m}^3

fall into the intervals

10 \le c < 20

and

20 \le c < 30

.

3

Calculate the conditional probability as a decimal.

\frac{95}{125} = 0.76

Dividing the favorable outcomes (

95

) by the total possible outcomes in the restricted sample space (

125

) yields

0.76

.

Key Concept

Conditional probability and sample space restriction in grouped frequency tables

Question 10Question →

The frequency distribution table below summarizes the calibration offset errors, $x$ (in microvolts, $\mu\text{V}$ ), measured for a sample of $200$ precision voltage sensors in a robotics laboratory.

Offset Error Interval ( $\mu\text{V}$ )	Frequency
$0 \le x < 10$	$32$
$10 \le x < 20$	$f_1$
$20 \le x < 30$	$68$
$30 \le x < 40$	$f_2$
$40 \le x < 50$	$24$

The estimated mean offset error calculated using the midpoints of the five class intervals is equal to $24.8\ \mu\text{V}$ . If a sensor is selected at random from among those with an offset error of at least $20\ \mu\text{V}$ , what is the probability that its offset error is less than $40\ \mu\text{V}$ ?

\frac{14}{17}

\frac{14}{25}

\frac{25}{31}

\frac{11}{34}

\frac{17}{25}

Show answer & explanation

Answer: $\frac{14}{17}$

Answer

The correct answer is

\frac{14}{17}

, which represents the conditional probability that a sensor's offset error is less than

40\ \mu\text{V}

given that it is at least

20\ \mu\text{V}

.

The correct answer is

\frac{14}{17}

. Solving the system of equations formed by the total sample size (

f_1 + f_2 = 76

) and the estimated midpoint mean (

3f_1 + 7f_2 = 404

) yields

f_1 = 32

and

f_2 = 44

. The number of sensors with offset

\ge 20\ \mu\text{V}

is

68 + 44 + 24 = 136

. Among these, the number of sensors with offset

< 40\ \mu\text{V}

is

68 + 44 = 112

. Thus, the conditional probability is

\frac{112}{136} = \frac{14}{17}

.

Step-by-Step Solution

1

Set up an equation for the total frequency of the sample.

32 + f_1 + 68 + f_2 + 24 = 200 \implies f_1 + f_2 = 76

The sum of all class frequencies must equal the given total sample size of 200.

2

Set up an equation for the estimated mean using interval midpoints.

Midpoints are

5, 15, 25, 35, 45

. Total weighted sum

= 32(5) + f_1(15) + 68(25) + f_2(35) + 24(45) = 160 + 15f_1 + 1700 + 35f_2 + 1080 = 2940 + 15f_1 + 35f_2

. Mean

= \frac{2940 + 15f_1 + 35f_2}{200} = 24.8 \implies 15f_1 + 35f_2 = 2020 \implies 3f_1 + 7f_2 = 404

.

The estimated mean of grouped data is the sum of products of interval midpoints and frequencies divided by total sample size.

3

Solve the linear system of equations for

f_1

and

f_2

.

Multiply

f_1 + f_2 = 76

by

3

to get

3f_1 + 3f_2 = 228

. Subtracting from

3f_1 + 7f_2 = 404

gives

4f_2 = 176 \implies f_2 = 44

. Then

f_1 = 76 - 44 = 32

.

Elimination yields exact unknown frequencies for the remaining intervals.

4

Determine the conditional sample space and target frequency.

Condition (offset

\ge 20\ \mu\text{V}

): Intervals

[20, 30), [30, 40), [40, 50)

with total frequency

68 + 44 + 24 = 136

. Target condition (offset

< 40\ \mu\text{V}

within condition): Intervals

[20, 30)

and

[30, 40)

with frequency

68 + 44 = 112

.

Conditional probability restricts the denominator to sensors meeting the given condition.

5

Calculate the final probability fraction.

Probability

= \frac{112}{136} = \frac{14}{17}

.

Dividing target count by conditional total count yields the simplified fraction.

Key Concept

Grouped Data Mean Estimation & Conditional Probability from Frequency Distributions

Estimated Time:2m 30s

Question 11Question →

A environmental monitoring group collected $200$ soil samples from a nature reserve and recorded their pH levels in the frequency table below:

pH Level Range	Number of Samples
$5.0 – 5.9$	$40$
$6.0 – 6.9$	$70$
$7.0 – 7.9$	$50$
$8.0 – 8.9$	$30$
$9.0 – 9.9$	$10$

What percent of the soil samples with a pH level of at least $6.0$ have a pH level in the range $6.0$ to $7.9$ ?

43.75\%

60.00\%

75.00\%

80.00\%

68.75\%

Show answer & explanation

Answer: $75.00\%$

Answer

75.00\%

To find the desired percentage, first restrict the sample space to all soil samples having a pH level of at least

6.0

. Adding the frequencies for

6.0–6.9

(

70

),

7.0–7.9

(

50

),

8.0–8.9

(

30

), and

9.0–9.9

(

10

) gives a base total of

160

samples. Next, find the number of samples within that group that fall in the range

6.0

to

7.9

, which is

70 + 50 = 120

. The percentage is calculated as

(120 / 160) \times 100\% = 75.00\%

.

Step-by-Step Solution

1

Determine the conditional base population (samples with pH

\ge 6.0

).

Sum of frequencies for ranges

6.0-6.9

,

7.0-7.9

,

8.0-8.9

, and

9.0-9.9

:

70 + 50 + 30 + 10 = 160

.

The question asks 'of the soil samples with a pH level of at least 6.0', which restricts the total base to these four intervals.

2

Determine the target frequency (samples with pH in range

6.0

to

7.9

).

Sum of frequencies for ranges

6.0-6.9

and

7.0-7.9

:

70 + 50 = 120

.

This captures all qualifying samples within the specified target interval.

3

Calculate the conditional percentage.

\frac{120}{160} \times 100\% = 0.75 \times 100\% = 75.00\%

.

Divide the target sub-group count by the conditional base count and convert to a percentage.

Key Concept

Conditional Relative Frequency in Grouped Data Tables

Estimated Time:1m 30s

Question 12Question →

A reliability study recorded the operating lifespan, $t$ (in thousands of hours), for a sample of $250$ semiconductor laser diodes. The results are summarized in the grouped frequency table below.

Lifespan $t$ (thousands of hours)	Frequency
$0 \le t < 4$	$35$
$4 \le t < 8$	$55$
$8 \le t < 12$	$80$
$12 \le t < 16$	$50$
$16 \le t < 20$	$30$

Laser diodes with an operating lifespan of at least $6{}000$ hours ( $t \ge 6$ ) but less than $14{}000$ hours ( $t < 14$ ) are designated as high-efficiency units. Assuming that the values within each class interval are uniformly distributed, what percentage of the $250$ laser diodes in the sample are designated as high-efficiency units?

Show answer & explanation

Answer: 53

Answer

53%

To find the percentage of diodes with lifespans between

6{}000

and

14{}000

hours (

6 \le t < 14

), evaluate the relevant intervals. For

4 \le t < 8

, the sub-interval

[6, 8)

represents

\frac{8-6}{8-4} = 0.5

of the interval width, containing

0.5 \times 55 = 27.5

diodes. The interval

[8, 12)

is fully contained, contributing

80

diodes. For

12 \le t < 16

, the sub-interval

[12, 14)

represents

\frac{14-12}{16-12} = 0.5

of the interval width, containing

0.5 \times 50 = 25

diodes. Totaling these gives

27.5 + 80 + 25 = 132.5

diodes. Expressed as a percentage of the total

250

diodes,

\frac{132.5}{250} \times 100\% = 53\%

.

Step-by-Step Solution

1

Determine the estimated number of diodes in the partial interval

6 \le t < 8

.

The target range

[6, 8)

covers half of the interval

[4, 8)

width of

4

units. With uniform distribution, the count is

0.5 \times 55 = 27.5

diodes.

Linear interpolation estimates frequencies for sub-intervals within grouped data.

2

Include the count for the complete interval

8 \le t < 12

.

All

80

diodes in this interval fall within

6 \le t < 14

.

The entire interval is fully contained within the target upper and lower bounds.

3

Determine the estimated number of diodes in the partial interval

12 \le t < 14

.

The target range

[12, 14)

covers half of the interval

[12, 16)

width of

4

units. With uniform distribution, the count is

0.5 \times 50 = 25

diodes.

Linear interpolation estimates frequencies for the upper partial boundary.

4

Sum the target diode counts and convert to a percentage of the total sample.

Total target count =

27.5 + 80 + 25 = 132.5

diodes. Percentage =

\frac{132.5}{250} \times 100\% = 53\%

.

Divide the calculated frequency sum by the total sample size of 250 and multiply by 100.

Key Concept

Grouped Frequency Distribution and Linear Interpolation

Estimated Time:2m 0s

Question 13Question →

A quality control manager at a pharmaceutical manufacturing facility evaluated the disintegration time, $t$ (in seconds), for a batch of $400$ coated tablets. The results are summarized in the frequency distribution table below:

Disintegration Time $t$ (seconds)	Frequency
$10 \le t < 20$	$40$
$20 \le t < 30$	$110$
$30 \le t < 40$	$160$
$40 \le t < 50$	$60$
$50 \le t < 60$	$30$

Which of the following statements regarding the distribution of disintegration times must be true? Select all that apply.

Select all that apply

The estimated mean disintegration time of the batch, calculated using class midpoints, is

33.25

seconds.

The median disintegration time of the batch lies within the interval

30 \le t < 40

.

Exactly

40\%

of the tablets in the batch have a disintegration time of less than

30

seconds.

The ratio of the number of tablets with a disintegration time of at least

30

seconds to the number of tablets with a disintegration time of less than

30

seconds is

5

to

3

.

More than

25\%

of the tablets in the batch have a disintegration time of

50

seconds or greater.

Show answer & explanation

Answer: The estimated mean disintegration time of the batch, calculated using class midpoints, is $33.25$ seconds.; The median disintegration time of the batch lies within the interval $30 \le t < 40$ .; The ratio of the number of tablets with a disintegration time of at least $30$ seconds to the number of tablets with a disintegration time of less than $30$ seconds is $5$ to $3$ .

Answer

The statements asserting that the estimated mean is

33.25

seconds, that the median lies in the interval

30 \le t < 40

, and that the ratio of tablets taking at least

30

seconds to those taking less than

30

seconds is

5

to

3

are all correct.

The estimated mean of

33.25

seconds is correctly computed from class midpoints

(13,300 / 400)

. The median interval

30 \le t < 40

correctly encompasses the middle values (

200^{\text{th}}

and

201^{\text{st}}

observations out of

400

). The part-to-part ratio of tablets with disintegration time

\ge 30

seconds (

250

) to

< 30

seconds (

150

) simplifies to

250:150 = 5:3

.

Step-by-Step Solution

1

Calculate the estimated mean using class midpoints.

Midpoints are

15, 25, 35, 45, 55

. Total product sum

\sum f \cdot m = (40 \times 15) + (110 \times 25) + (160 \times 35) + (60 \times 45) + (30 \times 55) = 13,300

. Mean

= 13,300 / 400 = 33.25

seconds.

The mean of grouped data is estimated by taking the weighted sum of interval midpoints divided by total sample size.

2

Determine the interval containing the sample median.

Cumulative frequencies:

10 \le t < 20 \rightarrow 40

;

20 \le t < 30 \rightarrow 150

;

30 \le t < 40 \rightarrow 310

. The

200^{\text{th}}

and

201^{\text{st}}

data points fall between cumulative counts

150

and

310

, putting the median in

30 \le t < 40

.

The median corresponds to the middle position (

N/2 = 200

) of ordered data.

3

Evaluate the percentage of tablets with disintegration time less than

30

seconds.

Count

= 40 + 110 = 150

. Percentage

= (150 / 400) \times 100\% = 37.5\%

.

Part-to-whole percentage requires dividing the sum of frequencies below

30

seconds by the total sample size

400

.

4

Compute the part-to-part ratio of tablets with

t \ge 30

seconds versus

t < 30

seconds.

Tablets with

t \ge 30

:

160 + 60 + 30 = 250

. Tablets with

t < 30

:

40 + 110 = 150

. Ratio

= 250 : 150 = 5 : 3

.

Formulating the part-to-part ratio comparing the upper three intervals to the lower two intervals.

5

Evaluate the percentage of tablets with disintegration time

t \ge 50

seconds.

Count

= 30

. Percentage

= (30 / 400) \times 100\% = 7.5\%

.

Checking if

7.5\%

exceeds

25\%

shows the statement is false.

Key Concept

Grouped Data Analysis: Mean Estimation, Median Interval Identification, and Relative Frequency Calculations

Frequency Distributions, Tables, and Grouped Data

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