Question

This week, a very large running race (5K) occured in Denver. The times were normally distributed, with a mean of 23.25 minutes and a standard deviation of 4.96 minutes.
Report your answers accurate to 2 decimals


a. What percent of runners took 15.36 minutes or less to complete the race?  %


b. What time in minutes is the cutoff for the fastest 5.4 %?  Minutes


c. What percent of runners took more than 18.9 minutes to complete the race?

Answer 1

Part A)

Values are given as,
Mean= 23.25,
Standard Deviation= 4.96,
x = 15.36,

We need to find P(x < 15.36)

Using below formula

Put the values in the formula,
P[(z < (15.36 - 23.25)/4.96]
= (z < -1.59)
z score is -1.59

Using z table, we get,
P(z < -1.59) = 0.0559

Hence, P(z < -1.59) = 0.0559*100= 5.59 %

Part B)

Values are given as,
Mean = 23.25,
Standard Deviation = 4.96,

z= -1.607 (value of 5.4th percentile)

Calculating raw-score:

x= 15.2781

x= 15.28

Part C)

Values are given as,
Mean = 23.25,
Standard Deviation = 4.96,
x = 18.9,

We need to find P(x > 18.9)

Using below formula


Put the values in the formula,
P[(z < (18.9 - 23.25)/4.96]
= (z < -0.88)
z score is -0.88

Using z table, we get,
P(z < -0.88) = 0.18943

Now P(z > -0.88) = 1 - P(z < -0.88)
Now P(z > -0.88) = 0.8106

Hence, P(z > -0.88) = 0.8106*100 = 81.06%

Answers :

A) P(x < 15.36)= 5.59 %

B) x= 15.28

C) P(x > 18.9) = 81.06%