Question

In: Statistics and Probability

The monthly expenditures on food by single adults living in one neighborhood of Los Angeles are...

The monthly expenditures on food by single adults living in one neighborhood of Los Angeles are normally distributed with a mean of $410 and a standard deviation of $75. Determine the percentage of samples of size 9 that have mean expenditures within $20 of the population mean expenditure of $410.

Expert Solution

Solution :

Given that,

mean = = 410

standard deviation = = 75

n = 9

= = 410

= / n = 75 / 9 = 25

P(390 < < 430)

= P[(390 - 410) / 25< ( - ) / < (430 - 410) / 25 )]

= P( - 0.8< Z < 0.8 )

= P(Z < 0.8 ) - P(Z < - 0.8)

Using z table,

= 0.7881 - 0.2119

= 0.5762

orchestra answered 2 years ago

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