Chapter 9:
Problem 1:
Grocery store receipts show that customer purchases have a skewed distribution with a
mean of $32 and a standard deviation of $20.
Suppose a random sample of 100 shoppers are chosen, and their grocery bills are
tabulated.
(A) Find the probability that the average shopping bill (of the 100 shoppers) will be more
than $35.
)35Pr( X
=
100/20
3235
Pr Z
= Pr(Z > 1.5) = 1-Pr(Z < 1.5) = 1 - .9332 = 0.0668.
(B) Find the probability that the average shopping bill (of the 100 shoppers) will be less
than $30.
)30Pr( X
=
100/20
3230
Pr Z
= Pr(Z < -1) = 0.1587.
(C) Find the probability that the average shopping bill (of the 100 shoppers) will be
between $31 and $39.
)3931Pr( X
=
100/20
3239
100/20
3231
Pr Z
= Pr(-0.5 < Z < 3.5) = Pr(Z < 3.5) – Pr(Z
< -0.5) = 0.9998 – 0.3085 = 0.6913.
Problem 2:
Information on a packet of seeds claims that the germination rate is 92%. A packet of
seeds contains 160 seeds. Assume each packet is a SRS from the population of interest.
(A) Let
p
ˆ
represent the proportion of seeds in a packet that germinate. Find the
approximate sampling distribution of
p
ˆ
.
160
)92.1(92.
,92.~
ˆNp
N(0.92, 0.021)
(B) Using your findings from part (A), what is the probability that less than 90% of seeds
in a packet germinate?
