Question

The heights of 8th grade boys are normally distributed with a mean of 50 inches and a standard deviation ofDraw and label a normal distribution curve below. 4 pts)2) pos) Approximately 99.7% of the heights fall betweenandb) pts) The middle 58% of heights fall betweenandps) What percentage of grade boys are between 55 and 50inches tall2 pts) What is the probability that an sgrade boy will beshorter than 55 inches

Answer 1

EXPLANATION

Let's see the facts:

Heights of 8th grade boys ---------------> Normally distributed

Mean= 50 inches

Standard deviation= 4 inches

d) Probability that boy shorter than 56 inches:

x=56

P(x<56)

We know that z-score=

`z=(x-\mu)/(\sigma)`

Replacing terms:

`z=(56-50)/(4)=1.5`

P-value from Z-table =

P(x<56) = 0.93319

Answer: The probability that an 8th grade boy will be shorter than 56 inches is 0.93319*100= 93.3%.

c) Percentage of 8th grade boys that are between 56 and and 60 inches tall is:

`z-score_(56)=(56-50)/(4)=(3)/(2)`

P-value from Z-table =

P(x<60) = 0.93319

`z-score_(56)=(60-50)/(4)=(5)/(2)`

P-value from Z-table =

P(x<60) = 0.99379

Then, subtracting both probabilities.

P(56

Answer: the percentage of 8th grade boys that are between 56 and 60 inches tall is 0.0606*100 = 6.06%.

a) We have that,

According to the Empirical Rule, we can assevere that 50+- 3σ = 50+- 3*4 ---->

50 + 12 = 62 inches

50 - 12 = 38 inches

About 99.7% of 8th grade boys height will fall between 38 inches and 62 inches.

Answer: Approximately 99.7% of the heights fall between 38 inches and 62 inches.

b) According to the Empirical Rule, we can assevere that around 68% of the data will fall within one standard deviation of the mean. As we already know, the standard deviation σ is a natural yardstick for any measurements that follow a normal distribution.

So, 50 +- σ = 50 +- 4 ---->

50 + σ = 50 + 4 = 54

50 - σ = 50 - 4 = 46

Answer: About 68% of 8th grade boys height will fall between 46 inches and 54 inches.