Question

A sample of students has shown their pulse rate to be a mean of 72.9 with a standard deviation of 12.3 at rest. You participate in a dance concert at the health fair and your pulse rate is measured at 95.9 at the end. Your best friend who also participated had a pulse rate of 100.45. What percent of those dancing would have pulse rates that exceed each of your rates? Whose pulse rate was more unusual, you or your friend?

Answer 1

Given,

= 72.9 , = 12.3

We convert this to standard normal as

P( X < x) = P( Z < x - / )

When you pulse rate = 95.9,

P( X > 95.9) = P( Z > 95.9 - 72.9 / 12.3)

= P( Z > 1.87)

= 1 - P( Z < 1.87)

= 1 - 0.9693

= 0.0307

For your friends pulse rate = 100.45

P( X > 100.45) = P( Z > 100.45 - 72.9 / 12.3)

= P( Z > 2.24)

= 1 - P( Z < 2.24)

= 1 - 0.9875

= 0.0125

Since probability of your and your friends pulse rate exceeding 72.9 are less than 0.05, both the

pulse rates are unusual.

Since your friends pulse rate is more away from the mean, that is lesser probability than yours,

Your friends pulse rate is more unusual.