In: Statistics and Probability

Salaries for teachers in a particular elementary school district are normally distributed with a mean of $46,000 and a standard deviation of $4,500. We randomly survey ten teachers from that district. (Round your answers to the nearest dollar.)

A) Find the 90^{th} percentile for an individual
teacher's salary.

B)Find the 90^{th} percentile for the average teacher's
salary.

Solution:

Given that,

mean = = 46,000

standard deviation = = 4,500

Using standard normal table,

a ) P(Z < z) = 90%

P(Z < z) = 0.90

P(Z < 1.28) = 0.27

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 4,500 + 46,000

x = 51760

The 90th percentile =51760

b )n = 10

= 46,000

= / n = 4,50010 =1423.0249

P(Z < z) = 90%

P(Z < z) = 0.90

P(Z < 1.28) = 0.27

z = 1.28

Using z-score formula,

= z * +

= 1.28 * 1423.0249 + 46,000

= 47821.47

The 90th percentile =47821

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