Question

In: Statistics and Probability

The mean salary of people living in a certain city is $37,500 with a standard deviation...

The mean salary of people living in a certain city is $37,500 with a standard deviation of $2,348. A sample of 62 people is selected at random from those living in the city.

Find the probability that the mean income of the sample is within $500 of the population mean.

Round your answer to 4 decimal places.

Expert Solution

Solution :

Given that,

mean = = 37500

standard deviation = = 2348

n = 62

= = 37500

= / n = 2348 / 62 = 298.20

37500 ± 500 = 37000, 38000

P(37000 < < 38000)

= P[(37000 - 37500) / 298.20 < ( - ) / < (38000 - 37500) / 298.20 )]

= P(-1.68 < Z < 1.68)

= P(Z < 1.68) - P(Z < -1.68)

Using z table,

= 0.9535 - 0.0465

= 0.9070

orchestra answered 1 year ago

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