According to records, the amount of precipitation in a certain city on a November day has a mean of 0.10 inches, with a standard deviation of 0.07 inches. What is the probability that the mean daily precipitation will be 0.107 inches or more for a random sample of 40 November days (taken over many years)

0.26369

Step-by-step explanation:

Given that:

Mean precipitation (m) = 0.10 inches

Standard deviation (σ) = 0.07 inches

Probability that mean daily precipitation will be 0.107 inches (x) or more

x ≥ 0.107

Sample size (n) = 40

First find the Zscore

Using the relation :

Zscore = (x - m) / (σ / √n)

Zscore = (0.107 - 0.10) / (0.07 / √40)

Zscore = 0.007 / 0.0110679

Zscore = 0.632

Hence ,

P(Z > 0.632) = 0.26369 (Z probability calculator)

the one below it is a function

a_n = a_1 + (n - 1)d

the first term is 6.

the common difference d = ?

when you find d, replace n with 7 and d with the common difference to find your answer.