Question

Let *x*_{1}, *x*_{2}, . . . ,
*x*_{100} denote the actual net weights (in pounds)
of 100 randomly selected bags of fertilizer. Suppose that the
weight of a randomly selected bag has a distribution with mean 75
lb and variance 1 lb^{2}. Let *x* be the sample mean
weight (*n* = 100).

(a) What is the probability that the sample mean is between
74.75 lb and 75.25 lb? (Round your answer to four decimal
places.)

*P*(74.75 ≤ *x* ≤ 75.25) =

(b) What is the probability that the sample mean is less than 75
lb?

Answer #1

Given,

= 75 , = 1

Using central limit theorem,

P( < x) = P( Z < x - / / sqrt(n) )

a)

P(74.75 <= <= 75.25) = P( <= 75.25) - P( <= 74.75)

= P( Z <= 75.25 - 75 / 1 / sqrt(100) ) - P( Z <= 74.75 - 75 / 1 / sqrt(100) )

= P( Z <= 2.5) - P( Z <= -2.5)

= 0.9938 - 0.0062

= **0.9876**

b)

P( < 75) = P( Z < 75 - 75 / 1 / sqrt(100) )

= P( Z < 0)

= **0.5**

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PLEASE EXPLAIN, I WILL GIVE A THUMBS UP FOR A NICE EXPLANATION.
THANK YOU IN ADVANCE :)

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