Question

Suppose the mean cost for a population of a 30-day supply of a generic drug is $46.58, with a population standard deviation of $4.84. A sample of 100 orders of 30-day supplies of generic drugs is selected randomly.

(a) What is the probability that the sample mean cost is less than $46? Answer to the nearest ten-thousandth

(b) What is the probability that the sample mean GPA is between $46 and $48?

(c) Within what limits will 85% of the sample mean costs occur?

Answer 1

a)

mu= 46.58

n= 100

sigma= 4.84

X= 46

Z=(X-mu)*sqrt(n)/sigma

=(46-46.58)*sqrt(100)/4.84

-1.1983

probability =P(Z<-1.1983)

=0.115390

b)

mu= 46.58

n= 100

sigma= 4.84

X1= 46

X2= 48

P(-1.1983<z<2.9333)

=0.9983-0.22539

=0.8829

c)

x?= 46.58

sigma= 4.84

n= 100

alpha=0.15 then Z(alpha/2)= 1.0364333

Margin of error E=Z(alpha/2)*sigma/sqrt(n)

=1.0364*4.84/sqrt(100)

=0.501633761

85% Confidence interval for population mean =(x?-E,x?+E)

rounded lower bound= 46.0784

rounded upper bound= 47.0816

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