A quality control manager at a pharmaceutical manufacturing facility evaluated the disinte… — GRE General Test

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.

The estimated mean disintegration time of the batch, calculated using class midpoints, is $33.25$ seconds.Cevap
The median disintegration time of the batch lies within the interval $30 \le t < 40$ .Cevap
C
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 $3$ .Cevap
E
More than $25\%$ of the tablets in the batch have a disintegration time of $50$ seconds or greater.

Cevap

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

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

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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.

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.

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

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.

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.

Anahtar Kavram

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

Soru

Cevap

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