Question

The average annual precipitation for a large Midwest City is 30.85 inches with a standard deviation of 3.6 inches. Assume the variable is normally distributed. a. Find the probability that a randomly selected month will have less than 30 inches. b. Find the probability that the mean of a random selection of 32 months will have a mean less than 30 inches.

Answer 1

Answer:

firstly 30.85. +3.6=add and find answer

then 32+30=add and find answers

then minus the both answers and then find the correct answer

Answer 2

Answer:

Probability that the mean of random selection of 32 months will have a mean less than 30 inches will be 0.0901.

Step-by-step explanation:

Given,

μ = 30.85

σ = 3.2

$v_n = \sqrt{32}$

To find probability that mean will be less than 30 -

$P(\bar X < 30) = P(\frac{(\bar X - \mu)}{\frac{\sigma}{v_n} } < \frac{30-30.85}{\frac{3.6}{\sqrt{32} } } )$

$P(Z < -1.34) = 0.0901$

By using Z-table we got probability of mean of random selection of 32 months will have mean less than 30 inches will be 0.0901.