Question

Heights of 10-year-olds, regardless of gender, closely follow a normal distribution with a mean of 55 inches and a standard deviation of 6 inches.(a) What is the probability that a randomly chosen 10 year old is shorter than 48 inches?(b) What is the probability that a randomly chosen 10 year old is between 60 and 65 inches?(c) If the tallest 10% of the class is considered "very tall", what is the height cutoff for "very tall"?(d) The height requirement for Batman the Ride at Six Flags Magic Mountain is 54 inches. What percent of 10 year olds cannot go on this ride?

Answer 1

Answer:

a) 12.10% probability that a randomly chosen 10 year old is shorter than 48 inches.

b) 15.58% probability that a randomly chosen 10 year old is between 60 and 65 inches.

c) The cutoff for very tall is 62.68 inches.

d) 43.25% of 10 year olds cannot go on this ride.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 55, \sigma = 6$

(a) What is the probability that a randomly chosen 10 year old is shorter than 48 inches?

This is the pvalue of Z when X = 48. So:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{48 - 55}{6}$

$Z = -1.17$

$Z = -1.17$ has a pvalue of 0.1210.

So there is a 12.10% probability that a randomly chosen 10 year old is shorter than 48 inches.

(b) What is the probability that a randomly chosen 10 year old is between 60 and 65 inches?

This is the value of Z when X = 65 subtracted by the pvalue of Z when X = 60.

X = 65

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{65 - 55}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

X = 60

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 55}{6}$

$Z = 0.83$

$Z = 0.83$ has a pvalue 0.7967

So there is a 0.9525-0.7967 = 0.1558 = 15.58% probability that a randomly chosen 10 year old is between 60 and 65 inches.

(c) If the tallest 10% of the class is considered "very tall", what is the height cutoff for "very tall"?

This is the value of X when Z has a pvalue of 0.9. So it is X when $Z = 1.28$

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 55}{6}$

$X - 55 = 1.28*6$

$X = 7.68+55$

$X = 62.68$

The cutoff for very tall is 62.68 inches.

(d) The height requirement for Batman the Ride at Six Flags Magic Mountain is 54 inches. What percent of 10 year olds cannot go on this ride?

This is the pvalue of Z when X = 54.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{54 - 55}{6}$

$Z = -0.17$

$Z = -0.17$ has a pvalue of 0.4325.

So 43.25% of 10 year olds cannot go on this ride.

Answer 2

A normal distribution has the shape of a bell curve.

The probability that a selected 10-year-old is shorter than 48 inches is 0.122

The probability that a selected 10-year-old is between 60 and 65 inches is 0.154

The cut-off height, if 10% is very tall is 62.70 inches

43.4% would make meet the height requirement of 54 inches

The mean and standard deviation heights are:

$\mu = 55$

$\sigma = 6$

(a) Probability of being shorter than 48 inches

This means that:

$x = 48$

Calculate the z score

$z = \frac{x - \mu}{\sigma}$

$z = \frac{48 - 55}{6}$

$z = \frac{-7}{6}$

$z = -1.167$

So, the probability is represented as:

$P(x < 48) = P(z <-1.167)$

Using the z table, we have:

$P(x < 48) = 0.122$

(b) Probability of being between 60 and 65 inches

This means that:

$x = 60, 65$

Calculate the z scores

$z = \frac{x - \mu}{\sigma}$

$z = \frac{60 - 55}{6} = 0.833$

$z = \frac{65 - 55}{6} = 1.667$

So, the probability is represented as:

$P(60<x<65) = P(z<1.667) - P(z<0.833)$

Using the z table, we have:

$P(60<x<65) = 0.952 - 0.798$

$P(60<x<65) = 0.154$

(c) Cut-off Height

We have:

$Tallest = 10\%$

The cut-off p value is:

$p = 1 - Tallest$

$p = 1 - 10\%$

$p = 1 - 0.10$

$p = 0.90$

The z value at $p = 0.90$ is:

$z = 1.282$

The cut-off height (x) is calculated as follows:

$z = \frac{x - \mu}{\sigma}$

$1.282= \frac{x - 55}{6}$

Cross multiply

$6 \times 1.282= x - 55}$

$7.692 = x - 55}$

Collect like terms

$x = 7.692 + 55$

$x = 62.692$

$x = 62.70$ --- approximated

(d) The percentage for height requirement of 54

Here, we have:

$z = 54$

Calculate the z score

$z = \frac{x - \mu}{\sigma}$

$z = \frac{54 - 55}{6}$

$z = \frac{ - 1}{6}$

$z = - 0.167$

The p score at $z = - 0.167$ is:

$p = 0.433685$

Convert to percentage

$p = 0.433685 \times 100\%$

$p = 43.3685\%$

Approximate

$p = 43.4\%$

Hence, 43.4% would make meet the height requirement

See attachment for the curve of the normal distribution

Read more about normal distributions at:

https://brainly.com/question/11876263