Normally distributed

Suppose the height of male youngsters is normally distributed with a mean of 60 inches and a standard deviation of 10. what percentage of the boy's height would we expect to be between 44 and 75, less than 49, and 76 and more?

Correct answer:

p₁ =  87.84 %
p₂ =  13.57 %
p₃ =  5.48 %

Step-by-step explanation:

$$\begin{gathered} m=60\text{in} \\ \sigma =10\text{in} \\ {p}_1=N(44<x<75) \\ {p}_1=100\cdot 0.8784=87.84\% \end{gathered}$$

Use the normal distribution table.

$$\begin{gathered} p_2=N(x<49) \\ {p}_2=100\cdot 0.1357=13.57\% \end{gathered}$$

Use the normal distribution table.

$$\begin{gathered} p_3=N(x>76) \\ {p}_3=100\cdot 0.0548=5.48\% \end{gathered}$$

Use the normal distribution table.

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