Question

The number of eggs that a female house fly lays during her lifetime is normally distributed with mean 840 and standard deviation 116. Random samples of size 82 are drawn from this population, and the mean of each sample is determined. What is the probability that the mean number of eggs laid would differ from 840 by less than 30? Round your answer to four decimal places.

If samples of size 39 are taken from a bimodal population, the distribution of sample means will be approximately normal. How can I be so sure of this?

A. The Law of Large Numbers says so
B. The Central Limit Theorem says so
C. The data is normal because the problem says so
D. It is a basic property of probability

Answer 1

Solution :

Given that ,

mean = = 840

standard deviation = = 116

n = 39

B. The Central Limit Theorem says so

=   = 840

= / n = 116 / 39 = 18.57

840 ± 30 = 810, 870

P(810 < < 870)  

= 1 - P[(810 - 840) / 18.57 < ( - ) / < (870 - 840) / 18.57)]

= 1 - P( -1.62 < Z < 1.62 )

= 1 - P(Z < 1.62) - P(Z < -1.62)

Using z table,  

= 1 - P( 0.9474 - 0.0526 )   

= 1 - 0.8948

= 0.1052