Question

In: Statistics and Probability

Assume the random variable X is normally distributed with mean =50 and standard deviation =7. Find...

Assume the random variable X is normally distributed with mean =50 and standard deviation =7. Find the 77 th percentile.

The mean incubation time of fertilized eggs is 2020 days. Suppose the incubation times are approximately normally distributed with a standard deviation of 11 day.

(a) Determine the 15th percentile for incubation times.

(b) Determine the incubation times that make up the middle 95%.

Solutions

Expert Solution

Question

Assume the random variable X is normally distributed with mean =50 and standard deviation =7. Find the 77 th percentile.

Solution:

The Z value for the 77th percentile by using z-table or excel is given as below:

Z = 0.738847

77th percentile = Mean + Z*SD

77th percentile = 50 + 0.738847*7

77th percentile = 55.17193

Question

The mean incubation time of fertilized eggs is 2020 days. Suppose the incubation times are approximately normally distributed with a standard deviation of 11 day.

(a) Determine the 15th percentile for incubation times.

The Z value for the 15th percentile by using z-table or excel is given as below:

Z = -1.03643

15th percentile = Mean + Z*SD

15th percentile = 2020 - 1.03643*11

15th percentile = 2008.599

(b) Determine the incubation times that make up the middle 95%.

The Z values for the middle 95% area by using z-table or excel are given as below:

Lower Z = -1.95996

Upper Z = 1.95996

Lower value = Mean + Z*SD = 2020 - 1.95996*11 = 1998.44

Upper value = Mean + Z*SD = 2020 + 1.95996*11 = 2041.56

Lower incubation time = 1998.44 days

Upper incubation time = 2041.56 days

orchestra answered 1 year ago

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