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
- Calculate the Z-score for 58.1 inches:
\[ Z = \frac{X - \mu}{\sigma} \]
Where:
- \( X = 58.1 , \text{inches} \)
- \( \mu = 53.4 , \text{inches} \)
- \( \sigma = 2.8 , \text{inches} \)
\[ Z = \frac{58.1 - 53.4}{2.8} = \frac{4.7}{2.8} \approx 1.6786 \]
- 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
- Calculate the Z-score for 54.2 inches:
\[ Z = \frac{54.2 - 53.4}{2.8} = \frac{0.8}{2.8} \approx 0.2857 \]
- Find the probability using the Z-score:
Using a Z-table or calculator, we find:
\[ P(Z < 0.2857) \approx 0.6134 \]
- 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
1 answer