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
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
Get Answers For Free
Most questions answered within 1 hours.