Question

Suppose that a principal of a local high school tracks the number of minutes his students spend texting on a given school day. He finds that the distribution of minutes spent texting is roughly normal with a mean of 60 and a standard deviation of 20. Use this information to answer the following questions.

1. Based on the statistics, find the two numbers of minutes that define the middle 95% of students in the distribution. What is the value for the lower number that you found?

2. Based on the statistics, find the two numbers of minutes that define the middle 95% of students in the distribution. What is the value for the higher number that you found?

Answer 1

Solution:-

Given that,

mean = = 60

standard deviation = = 20

Using standard normal table,

P( -z < Z < z) = 95%

= P(Z < z) - P(Z <-z ) = 0.95

= 2P(Z < z) - 1 = 0.95

= 2P(Z < z) = 1 + 0.95

= P(Z < z) = 1.95 / 2

= P(Z < z) = 0.975

= P(Z < 1.96 ) = 0.975

= z  ± 1.96

1) Using z-score formula,

x = z * +

x = -1.96 * 20 + 60

x = 20.8

lower value = 20.8

Using z-score formula,

x = z * +

x = 1.96 * 20 + 60

x = 99.2

higher value = 99.2