The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8.9 cm.
a. Find the probability that an individual distance is greater than 215.50 cm.
b. Find the probability that the mean for 15 randomly selected distances is greater than 204.00 cm.
c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?
Given mean = 205.5
Standard deviation = 8.9
Population is normally distributed, so we can use standard normal z table to estimate the proability
A)
P(x>215.5)
First we need to find the z score
Z = (x-mean)/s.d
(215.5-205.5)/8.9
Z = 1.12
From z table, p(z>1.12) = 0.1314
Therefore, the required probability is 0.1314
B)
P(x>204)
N = 15
Z = (x-mean)/(s.d/√n)
As standard deviation for sample is standard deviation/√n)
Z = (204-205.5)/(8.9/√15)
Z = -0.65
From z table, p(z>-0.65) = 0.7422
C)
We know that sample of any size from the population which is normally distributed is also normally distributed
As here given population is normally distributed, therefore given sample.of 15 is also normally distributed.
The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm...
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