Suppose a random variable has population mean 70 and populat

Suppose a random variable has population mean -70 and population standard deviation 25. What is the upper value of the probability interval containing 77% of the sample means of sample size n = 40? (Round to 1 decimal place.)

Suppose a random variable has population mean 80 and population standard deviation 18. What percentage of all possible sample means of sample size n = 36 will lie between 74 and 86? (Round to the nearest integer.)

Solution

Given a=1-0.77 = 0.23, Z(0.23) = 0.74 (from standard normal table)

So the upper bound is

xbar + Z*s/vn =70 +0.74*25/sqrt(40) =72.9

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P(74<xbar<86) = P((74-80)/(18/sqrt(36)) <(xbar-mean)/(s/vn) < (86-80)/(18/sqrt(36)))

=P(-2<Z<2) =0.9545 (from standard normal table)

i.e. 95%