Question

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?

Answer 1

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)

Answer 2

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.