Question

The manufacturer of cans of salmon that are supposed to have a net weight of 6 ounces tells you that the net weight is actually a normal random variable with a mean of 6.02 ounces and a standard deviation of 0.15 ounce. Suppose that you draw a random sample of 28 cans. Find the probability that the mean weight of the sample is less than 6.01 ounces.

Answer 1

Answer:

0.3605 is the required probability.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 6.02 ounces

Standard Deviation, σ = 0.15 ounce

Sample size, n = 28

We are given that the distribution of weight of can is a bell shaped distribution that is a normal distribution.

Formula:

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

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.15}{\sqrt{28}} = 0.028$

P(weight of the sample is less than 6.01 ounces)

$P( x < 6.01) = P( z < \displaystyle\frac{6.01 - 6.02}{0.028}) = P(z < -0.3571)$

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

$P(x < 6.01) =0.3605= 36.05\%$

0.3605 is the probability  that the mean weight of the sample is less than 6.01 ounces.