Question

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

Answer 1

Answer:

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 2

Using the normal distribution, it is found that:

• The cut off for the fastest 51.99% of times is of 24.09 minutes.

• 0.25% of runners took more than 32.128 minutes to complete the race.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem, we have that the mean and the standard deviation are, respectively, $\mu = 23.92, \sigma = 3.4$.

The cut off for the fastest 51.99% is the lowest 51.99% of times, that is, the 51.99th percentile, which is X when Z = 0.05.

$Z = \frac{X - \mu}{\sigma}$

$0.05 = \frac{X - 23.92}{3.4}$

$X - 23.92 = 0.05(3.4)$

$X = 24.09$

The cut off for the fastest 51.99% of times is of 24.09 minutes.

The proportion of runners who took more than 32.128 minutes to complete the race is 1 subtracted by the p-value of Z when X = 32.128, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{32.198 - 23.92}{3.4}$

$Z = 2.43$

$Z = 2.43$ has a p-value of 0.9925.

1 - 0.9925 = 0.0025.

0.0025 = 0.25% of runners took more than 32.128 minutes to complete the race.

More can be learned about the normal distribution at https://brainly.com/question/24663213