Question

Suppose that prices of women’s athletic shoes have a mean of $75 and a standard deviation of $17.89. What is the probability that the mean price of a random sample of 50 pairs of women’s athletic shoes will differ from the population mean by less than $5.00?

In order to estimate the number of calls to expect at a new suicide hotline, volunteers contact a random sample of 40 similar hotlines across the nation and find that the sample mean is 42.0 calls per month. Construct a 95% confidence interval for the mean number of calls per month. Assume that the population standard deviation is known to be 6.5 calls per month.

Answer 1

Solution :

1) Given that ,

mean = = 75

standard deviation = = 17.89

n = 50

=   = 75

= / n = 17.89 / 50 =

75 ± 5 = 70, 80

P(70 < < 80)  

= P[(70 - 75) / 2.53 < ( - ) / < (80 - 75) / 2.53)]

= P( -1.98 < Z < 1.98)

= P(Z < 1.98) - P(Z < -1.98)

Using z table,  

= 0.9761 - 0.0239

= 0.9522

2) Given that,

Z/2 = Z0.025 = 1.96

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

= 1.96 * ( 6.5 /  40)

= 2.01

At 95% confidence interval estimate of the population mean is,

  ± E

= 42.0  ± 2.01

= ( 39.99, 44.01 )