Question

The average diameter of sand dollars on a certain island is 4.00 centimeters with a standard deviation of 0.60 centimeters. If 16 sand dollars are chosen at random for a collection, find the probability that the average diameter of those sand dollars is more than 3.85 centimeters. Assume that the variable is normally distributed.

Answer 1

0.8413 is the required probability.

Explanation:

We are given the following information in the question:

Mean, μ = 4.00 centimeters

Standard Deviation, σ = 0.60 centimeters

Sample size, n = 16

We are given that the distribution of average diameter is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

Standard error due to sampling =

`=(\sigma)/(√(n)) = (0.60)/(√(16)) = 0.15`

P(diameter of sample is more than 3.85 centimeter)

P(x > 3.85)

`P( x > 3.85) = P( z > \displaystyle(3.85 - 4)/(0.15)) = P(z > -1)`

`= 1 - P(z \leq -1)`

Calculation the value from standard normal z table, we have,

`P(x > 3.85) = 1 -0.1587 = 0.8413`

0.8413 is the probability that the the average diameter of sample of 16 sand dollars is more than 3.85 centimeters.