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

Answer:

0.8413 is the required probability.

Step-by-step 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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.60}{\sqrt{16}} = 0.15$

P(diameter of sample is more than 3.85 centimeter)

P(x > 3.85)

$P( x > 3.85) = P( z > \displaystyle\frac{3.85 - 4}{0.15}) = P(z > -1)$

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

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.