-12 points POOStat5 BE.033. Let x1, x2, . . . , x100 denote the actual net weigh

Question

-12 points POOStat5 BE.033. 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 a distribution with mean 35 lb and variance 1 lb2. Let x be the sample mean weight (n 100). Notes Ask Y (a) What is the probability that the sample mean is between 34.65 lb and 35.35 b? (Round your answer to four decimal places.) A34 .655 x 35.35) = (b) What is the probability that the sample mean is greater than 35 lb? You may need to use the appropriate table in Appendix A to answer this question.

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 35
standard Deviation ( sd )= 1/ Sqrt ( 100 ) =0.1
sample size (n) = 100
a.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 34.65) = (34.65-35)/1/ Sqrt ( 100 )
= -0.35/0.1
= -3.5
= P ( Z <-3.5) From Standard Normal Table
= 0.00023
P(X < 35.35) = (35.35-35)/1/ Sqrt ( 100 )
= 0.35/0.1 = 3.5
= P ( Z <3.5) From Standard Normal Table
= 0.99977
P(34.65 < X < 35.35) = 0.99977-0.00023 = 0.99953
b.
P(X > 35) = (35-35)/1/ Sqrt ( 100 )
= 0/0.1= 0
= P ( Z >0) From Standard Normal Table
= 0.5

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