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Sagot :
Answer:
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%
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