Question

I need help with my homewrok please check work when done

Answer 1

Given that;

The amount of 5th and 6th graders is normally distributed with a mean of 4 inches and standard deviation of 1 inch.

To find the percentage of 5th and 6th graders that grows between 1 and 3 inches, we will apply the z score formula

`\begin{gathered} z=(x-\mu)/(\sigma) \\ Where \\ x\text{ is the observed value} \\ \mu\text{ is the mean} \\ \sigma\text{ is the standard deviation} \end{gathered}`

Where, x = 1 inch

`\begin{gathered} z=(x-\mu)/(\sigma) \\ z=(1-4)/(1)=(-3)/(1)=-3 \\ z=-3 \end{gathered}`

Where, x = 3 inches

`\begin{gathered} z=(x-\mu)/(\sigma) \\ z=(3-4)/(1)=(-1)/(1)=-1 \\ z=-1 \end{gathered}`

Using the z score table,

`P(-3For the 5th graders, the percentage is[tex]=0.15731*100\%=15.731\%`

For the 6th graders, the percentage is the same as the percentage of the 5th graders.

The percentage of the 6th graders is

`15.731\%`

The percentage of 5th and 6th graders that grows between 1 and 3 inches will be

Since, they have different peaks, it is not appropriate to use normal distribution.

Hence, the answer is A