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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.

User Fady Kamal
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1 Answer

6 votes

Answer:

0.3605 is the required probability.

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(x-\mu)/(\sigma)

Standard error due to sampling =


=(\sigma)/(√(n)) = (0.15)/(√(28)) = 0.028

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


P( x < 6.01) = P( z < \displaystyle(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.

User Holger Frohloff
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