Question
Suppose that prices of women's athletic shoes have a mean of $75.15 and a standard deviation...
Suppose that prices of women's athletic shoes have a mean of
$75.15 and a standard deviation of $17.89. What is the z-score of
the mean
price $80.15 from a random sample of 50 pairs of women's athletic
shoes
Homework Answers
Answer #1
Solution :
Given that,
mean = 
= $75.15
standard deviation = 
=$17.89
n = 50
= $75.15
=
/ 
n = 17.89 /
50 = 2.5300
= $80.15
Z =
-
/
= 80.15 - 75.15 / 2.5300 = 1.9763
z-score = 1.9763
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