Question

the mean time taken by all participants to run a road race was found to be 195 minutes with a standard deviation of 14 minutes. what interval contains at leat

1. 84 %

2. 93.85 of the data values.

Answer 1

Solution :

Given that ,

mean = = 195 minutes

standard deviation = = 14 minutes

1) Using standard normal table,

P( -z < Z < z) = 84%

= P(Z < z) - P(Z <-z ) = 0.84

= 2P(Z < z) - 1 = 0.84

= 2P(Z < z) = 1 + 0.84

= P(Z < z) = 1.84 / 2

= P(Z < z) = 0.92

= P(Z < 1.41) = 0.92

= z  ± 1.41

Using z-score formula,

x = z * +

x = -1.41 * 14 + 195

x = 175.26 minutes

Using z-score formula,

x = z * +

x = 1.41 * 14 + 195

x = 214.74 minutes

The 84% interval is 175.26 to 214.74 minutes.

2) Using standard normal table,

P( -z < Z < z) = 93.85%

= P(Z < z) - P(Z <-z ) = 0.9385

= 2P(Z < z) - 1 = 0.9385

= 2P(Z < z) = 1 + 0.9385

= P(Z < z) = 1.9385 / 2

= P(Z < z) = 0.9693

= P(Z < 1.41) = 0.9693

= z  ± 1.87

Using z-score formula,

x = z * +

x = -1.87 * 14 + 195

x = 168.82 minutes

Using z-score formula,

x = z * +

x = 1.87 * 14 + 195

x = 221.18 minutes

The 93.85% interval is 168.82 to 221.18 minutes.