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A group of 80 frogs was observed. The mean distance of their hops is 69 inches with a standard deviation of 3.5 inches. How many frogs would you expect to jump more than 72.5 inches?

User Occhiso
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2 Answers

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Hello,


z=(X-69)/(3.5) \\\\For\ X=72.5, \\\\z=(72.5-69)/(3.5) =1\\

Using table of a normal reduced law:

p(z≤1)=0.8413

Thus p(z≥1)=1-0.8413=0.1587

There are 80*0.1587=12.696 ≈13 (frogs)

User Hitesh Singh
by
7.4k points
3 votes

Answer:

12 frogs

Explanation:

Hello,

Using table of a normal reduced law:

p(z≤1)=0.8413

Thus p(z≥1)=1-0.8413=0.1587

There are 80*0.1587=12.696 ≈12 (frogs) you don't round up because you cant have .7 percent of a frog.

ps. I copy and pasted caylus's response but corrected their answer because it was correct except the rounding up part.

User Paul Calcraft
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7.6k points
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