In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed with a mean of 53.4 inches, and standard deviation of 2.8 inches.

Question

A) What is the probability that a randomly chosen child has a height of less than 58.1 inches?

Answer=___
(Round your answer to 3 decimal places.)

B) What is the probability that a randomly chosen child has a height of more than 54.2 inches?

Answer=___
(Round your answer to 3 decimal places.)

Answer 1

To solve these problems, we'll use the properties of the normal distribution. The given mean height ($\mu$) is 53.4 inches, and the standard deviation ($\sigma$) is 2.8 inches. We will use the Z-score formula to convert the heights into Z-scores, which can then be used to find probabilities from standard normal distribution tables or a calculator.

A) Probability that a randomly chosen child has a height of less than 58.1 inches

1. Calculate the Z-score for 58.1 inches:

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

Where:

• $X=58.1,\text{inches}$
• $\mu =53.4,\text{inches}$
• $\sigma =2.8,\text{inches}$

$$Z=\frac{58.1-53.4}{2.8}=\frac{4.7}{2.8}\approx 1.6786$$

2. Look up the Z-score in the standard normal distribution table or use a calculator to find the probability:

Using a Z-table or calculator, we find:

$$P(Z<1.6786)\approx 0.9535$$

So, the probability that a randomly chosen child has a height of less than 58.1 inches is approximately:

$$\text{Answer A}=0.954\text{ (rounded to three decimal places)}$$

B) Probability that a randomly chosen child has a height of more than 54.2 inches

1. Calculate the Z-score for 54.2 inches:

$$Z=\frac{54.2-53.4}{2.8}=\frac{0.8}{2.8}\approx 0.2857$$

2. Find the probability using the Z-score:

Using a Z-table or calculator, we find:

$$P(Z<0.2857)\approx 0.6134$$

3. Since we want the probability of a height greater than 54.2 inches:

$$P(Z>0.2857)=1-P(Z<0.2857)$$

$$P(Z>0.2857)\approx 1-0.6134\approx 0.3866$$

So, the probability that a randomly chosen child has a height of more than 54.2 inches is approximately:

$$\text{Answer B}=0.387\text{ (rounded to three decimal places)}$$

Final Answers:

• A) 0.954
• B) 0.387