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## Sagot :

**Answer:**

A. 637

B. 0.2240 (4 dp) = 22.4% (nearest tenth)

**Step-by-step explanation:**

__Normal Distribution__

[tex]\sf X \sim N(\mu, \sigma^2)[/tex]

Given:

- [tex]\sf \mu = 190.6[/tex]

- [tex]\sf \sigma=5.8[/tex]

[tex]\implies \sf X \sim N(190.6, 5.8^2)[/tex]

__Part A__

[tex]\begin{aligned}\sf P(180 < X < 190) & = \sf P(X < 190)-P(X\leq 180)\\& = \sf 0.4588035995-0.03380583874\\& = \sf 0.4249977608\end{aligned}[/tex]

Total number of bodybuilders = 1500

Therefore, the number of bodybuilders between 180 and 190 pounds is:

[tex]\begin{aligned}\sf P(180 < X < 190) \cdot1500 & = \sf 0.4249977608 \cdot 1500\\& = \sf 637.4966412\\& = \sf 637\end{aligned}[/tex]

__Part B__

[tex]\begin{aligned}\sf P(X > 195) & = \sf 1-P(X\leq 195)\\& = \sf 1-0.7759602537\\ & = \sf 0.2240397463\\ & = \sf 0.2240\:(4\:dp)\end{aligned}[/tex]