This week a very large running race 5K occurred in Denver the times were normally distributed with a mean of 23.92 minutes and the standard deviation of 3.4 minutes what time in minutes is the cut off for the fastest 51.99% and what percent of runners took more than 32.128 minutes to complete the race

a) 24.09 minutes

b) 0.8%

Step-by-step explanation:

This week a very large running race 5K occurred in Denver the times were normally distributed with a mean of 23.92 minutes and the standard deviation of 3.4 minutes what time in minutes is the cut off for the fastest 51.99%

Z score for 51.99% = 0.05

z = (x-μ)/σ,

where x is the raw score = ? minutes

μ is the population mean = 23.92 minutes

σ is the population standard deviation = 3.4 minutes

0.05 = x - 23.92/3.4

Cross Multiply

0.05 × 3.4= x - 23.92

0.17 = x - 23.92

0.17 + 23.92 = x

x = 24.09 minutes

What percent of runners took more than 32.128 minutes to complete the race

= z score

z = (x-μ)/σ,

where x is the raw score = 32.128 minutes

μ is the population mean = 23.92 minutes

σ is the population standard deviation = 3.4 minutes

z = 32.128 - 23.92/3.4

z = 2.41412

P-value from Z-Table:

P(x<32.128) = 0.99211

P(x>32.128) = 1 - P(x<32.128) = 0.0078867

Approximately =0.008

Converting to percentage = 0.008 × 100

= 0.8%

answer: the is put the 450

step-by-step explanation: