This one is tricky! Imagine that you meet a new friend who is also a beginner, and she can run the 5k in 23.5 minutes. You wonder what percentage of the beginner running population could run the 5k faster than your new friend (that is, what percentage of the population has a time that is less than your new friend

38.74%

Step-by-step explanation:

Given the data:

21 21 22 22 23 23 23 24 24 24 24 24 25 25 25 26 26 27 27

We obtain the beginner running population and standard deviation

Population mean, μ = Σx/n = 456/19 = 24

Standard deviation, σ = 1.747 (using calculator)

Friend's Runtime, x = 23.5 minutes

Obtaining the friend's Zscore :

Z = (x - μ) / σ

Z = (23.5 - 24) / 1.747

Z = - 0.286

Obtaining the Pvalue :

Using a standard normal distribution table :

P(Z < - 0.286) = 0.38744

Hence. Percentage of population that has lesser time :

0.38744 * 100% = 38.74%

x= 149. 24+7=31. 180-31=149

your welcome

they disappear if your absent from to long. i found out that the hard way.

step-by-step explanation:

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