Question

According to the Bureau of Labor Statistics, the annual wage for a federal employee in 2017 was $76,250. Assume the population standard deviation is $28,025. A random sample of 34 federal employees is selected.

1. What is the probability that the sample mean will be less than $70,000?
2. What is the probability that the sample mean will be more than $74,000?
3. What is the probability that the sample mean will be between $76,000 and $81,000.

Answer 1

Solution :

Given that ,

mean = = $76250

standard deviation = = $28025

= / n = 28025 / 34 = 4806.2478

- P( <$70000 ) = P(( - ) / < (70000 - 76250) / 4806.2478)   

P(z < -1.30)

= 0.0968

= Probability = $0.0968

- P( > $74000) = 1 - P( < 74000)

= 1 - P[( - ) / < (74000 - 76250) / 4806.2478]

= 1 - P(z < -0.47)

= 1 - 0.3192 = 0.6808

= Probabilty = $0.6808

-P[(76000 - 76250) / 4806.2478< ( - ) / < (81000 - 76250) / 4806.2478)]

P(-0.05 < Z < 0.99)

= P(Z < 0.99) - P(Z < -0.05)   

= 0.8389 - 0.4801 = 0.3588

= Probabilty = $0.3588