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3 years ago

15

A produce supplier ships boxes of produce to individual customers. The distribution of weights of shipped boxes is approximately

normal with mean 36 pounds and standard deviation 4 pounds. Which expression represents the weight, in pounds, at the 75th percentile of the distribution?
(A) -1.96(4)+ 36
(B) -0.25(4) + 36
(C) 0.25(4)+ 36
(D) 0.67(4)+ 36
(E) 0.75(4)+ 36

2 answers:

Romashka [77]3 years ago

6 0

Answer:

Option D.

Step-by-step explanation:

Given information: The distribution of weights of shipped boxes is approximately normal.

Population mean = 36

Standard deviation = 4

We need to find the 75th percentile of the distribution

75% = 0.75

From z-table it is clear that the value of z at 0.75 is 0.674.

75th percentile = Mean + z value at 75%(Standard deviation)

75th percentile = 36 + 0.674(4)

In can be rewritten as

75th percentile = 0.67(4) + 36

Therefore, the correct option is D.

schepotkina [342]3 years ago

5 0

Answer:

$0.67(4)+ 36$

Option D

Step-by-step explanation:

Given that a producer ships boxes of produce to individual customers.(say x)

X is a normal random variable with mean = 36 and std dev =4 lbs

By definition of std normal variate we know that

$\frac{x-36}{4}$ is N(0,1)

75th percentile of std normal distribution is

0.675

Hence corresponding x would be

$36+0.675(4)$

Option D matches with this value

Hence answer is option D

$0.67(4)+ 36$

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