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In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally di...Question
In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed with a mean of 57 inches, and standard deviation of 5.2 inches.
What is the probability that the height of a randomly chosen child is between 56.3 and 66.4 inches? Do not round until you get your your final answer, and then round to 3 decimal places.
Answer=
(Round your answer to 3 decimal places.)
What is the probability that the height of a randomly chosen child is between 56.3 and 66.4 inches? Do not round until you get your your final answer, and then round to 3 decimal places.
Answer=
(Round your answer to 3 decimal places.)
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Answered by
GPT-4o mini
To find the probability that the height of a randomly chosen ten-year-old child in Heightlandia is between 56.3 and 66.4 inches, we can use the properties of the normal distribution. We will first determine the z-scores for both height values and then use the standard normal distribution to find the probabilities.
The z-score formula is:
\[
z = \frac{(X - \mu)}{\sigma}
\]
where:
- \(X\) is the value for which we are calculating the z-score,
- \(\mu\) is the mean,
- \(\sigma\) is the standard deviation.
Given:
- Mean (\(\mu\)) = 57 inches
- Standard deviation (\(\sigma\)) = 5.2 inches
### Step 1: Calculate the z-scores
**For 56.3 inches:**
\[
z_1 = \frac{(56.3 - 57)}{5.2} = \frac{-0.7}{5.2} \approx -0.1346153846
\]
**For 66.4 inches:**
\[
z_2 = \frac{(66.4 - 57)}{5.2} = \frac{9.4}{5.2} \approx 1.8076923077
\]
### Step 2: Look up the z-scores in the standard normal distribution table
Using a standard normal distribution table or a calculator, we can find the probabilities associated with these z-scores.
**For \(z_1 \approx -0.135\):**
The cumulative probability (area to the left) for \(z_1\) is approximately:
\[ P(Z < -0.135) \approx 0.4477 \]
**For \(z_2 \approx 1.808\):**
The cumulative probability for \(z_2\) is approximately:
\[ P(Z < 1.808) \approx 0.9643 \]
### Step 3: Calculate the probability between the two z-scores
The probability that a child's height is between 56.3 inches and 66.4 inches is calculated by:
\[
P(56.3 < X < 66.4) = P(Z < z_2) - P(Z < z_1)
\]
\[
= P(Z < 1.808) - P(Z < -0.135) = 0.9643 - 0.4477 = 0.5166
\]
### Final Answer
Therefore, the probability that the height of a randomly chosen child is between 56.3 and 66.4 inches is approximately 0.5166. Rounding to three decimal places, we get:
\[
\text{Probability} \approx 0.517
\]
So, the final round answer is:
**Answer: 0.517**
The z-score formula is:
\[
z = \frac{(X - \mu)}{\sigma}
\]
where:
- \(X\) is the value for which we are calculating the z-score,
- \(\mu\) is the mean,
- \(\sigma\) is the standard deviation.
Given:
- Mean (\(\mu\)) = 57 inches
- Standard deviation (\(\sigma\)) = 5.2 inches
### Step 1: Calculate the z-scores
**For 56.3 inches:**
\[
z_1 = \frac{(56.3 - 57)}{5.2} = \frac{-0.7}{5.2} \approx -0.1346153846
\]
**For 66.4 inches:**
\[
z_2 = \frac{(66.4 - 57)}{5.2} = \frac{9.4}{5.2} \approx 1.8076923077
\]
### Step 2: Look up the z-scores in the standard normal distribution table
Using a standard normal distribution table or a calculator, we can find the probabilities associated with these z-scores.
**For \(z_1 \approx -0.135\):**
The cumulative probability (area to the left) for \(z_1\) is approximately:
\[ P(Z < -0.135) \approx 0.4477 \]
**For \(z_2 \approx 1.808\):**
The cumulative probability for \(z_2\) is approximately:
\[ P(Z < 1.808) \approx 0.9643 \]
### Step 3: Calculate the probability between the two z-scores
The probability that a child's height is between 56.3 inches and 66.4 inches is calculated by:
\[
P(56.3 < X < 66.4) = P(Z < z_2) - P(Z < z_1)
\]
\[
= P(Z < 1.808) - P(Z < -0.135) = 0.9643 - 0.4477 = 0.5166
\]
### Final Answer
Therefore, the probability that the height of a randomly chosen child is between 56.3 and 66.4 inches is approximately 0.5166. Rounding to three decimal places, we get:
\[
\text{Probability} \approx 0.517
\]
So, the final round answer is:
**Answer: 0.517**
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