Question

Cell phone cost is normally distributed with a mean of $83, and a standard deviation of $16. For a randomly selected cell phone, find the following:

What is the probability that the cost is more than $ 110?

What is the probability that the cost is between $ 90 and $100?

What is the cost of the top 2% of users?

Do problem 1 for a sample of 10 cell phones.

Answer 1

Solution :

Given that ,

1) mean = = $83

standard deviation = = $16

- = / n = 16 / 10 = 5.0596

- P( > $110) = 1 - P( < 110)

= 1 - P[( - ) / < (110 - 83) / 5.0596]

= 1 - P(z < 5.34)

= 1 - 1 = 0

= Probability = $0.0000

- = P[(90 - 83) / 5.0596< ( - ) / < (100 - 83) / 5.0596)]

= P(1.38 < Z < 3.36)

= P(Z < 3.36) - P(Z < 1.38)

= 0.9996 - 0.9162

= 0.0834

Probability = $0.0834

- Using standard normal table ,

P(Z > z) = 2%

1 - P(Z < z) = 0.02

P(Z < z) = 1 - 0.02 = 0.98

P(Z < 2.05) = 0.98

z = 2.05

Using z-score formula,

x = z * +

x = 2.05 * 5.0596 + 83 = 93.37

cost of the top 2% of users $93.37