Question

The distribution of the number of moths captured per night by a certain moth trap is approximately normal with mean 103. If 28 percent of the captures fall below 76 per night, which of the following equations can be used to find σσ, the standard deviation of the distribution?
A) 0.28= 103-76/ σ
B) 0.28= 76- 103/ σ
C) -0.58= 103-76/ σ
D) -0.58= 103-76/σ
E) 0.58= 76-103/ σ

Answer 1

Answer:

0.28 = (76 - 103)/σ (B)

Step-by-step explanation:

Let X be a random variable for capturing moths every night

Mean(μ) = 103

σ is the standard deviation

Pr(X<76) = 28%

Pr(X<76) = 0.28

For normal distribution, Z = (X - μ)/σ

Pr(X<76) = (X - μ)/σ

0.28 = (76 - 103)/σ

The equation to find the standard deviation is

0.28 = (76 - 103)/σ

Answer 2

Answer:C)   $0.58=\dfrac[103-76}{\sigma}$

Step-by-step explanation:

Given : The distribution of the number of moths captured per night by a certain moth trap is approximately normal with  $\mu=103$.

Also,  28 percent of the captures fall below 76 per night, which of the following equation,

On z-table , the z-value corresponding to the p-value of 28% is -0.58.

Since formula for z-score : $z=\dfrac{x-\mu}{\sigma}$ , where x is random variable and $\sigma$ is standard deviation.

For the current situation, the formula becomes with values  $\mu=103, x=76 , \&\ z=-0.58$ as

$-0.58=\dfrac{76-103}{\sigma}\\\\0.58=\dfrac[103-76}{\sigma}$

Hence, the equations can be used to find σσ, the standard deviation of the distribution :-

$0.58=\dfrac[103-76}{\sigma}$