Answer to Question #115809 in Statistics and Probability for Seifu

Question

The mean expenditure per customer at a department store is Birr 85 with standard deviation of Birr 9. If a random sample of 40 customers is taken, what is the probability that the sample average expenditure per customer will be
a) Birr 87 or more
b) Less than Birr 82
c) Between Birr 83 and 85

Answer 1

n=40, sd=9, sample mean =85

$z= \frac {x-\mu} {\frac {s} {\sqrt n}}$

a) $P(X \geq 87)$

=1-P(X<87)

P(X<87) =$\frac {87-85}{\frac {9}{\sqrt {40}}}$ =1.41

$\Phi (1.41)$ = 0.92073 from the tables

=1-0.92073

=0.07927

b) P(X<82)

=$=\frac {82-85}{\frac {9}{\sqrt {40}}}$ = - 2.11

$\Phi (-2.11)$ = 0.01786 from the standard normal table

=0.01786

c) P(83<X<85)

$Z_1= \frac {83-85}{\frac {9}{\sqrt {40}}}$ = - 1.41

$Z_2= \frac {85-85}{\frac {9}{\sqrt {40}}}$ =0

$\Phi (0) - \Phi (-1.41)$

=0.5 -0.07927

=0.42073