Question

# Let x1, x2, . . . , x100 denote the actual net weights (in pounds) of...

Let x1, x2, . . . , x100 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 lb2. 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?

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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