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A manufacturer knows that their items have a normally distributed length, with a mean of 7.3 inches, and standard deviation of...Question
A manufacturer knows that their items have a normally distributed length, with a mean of 16.9 inches, and standard deviation of 3.7 inches.
If one item is chosen at random, what is the probability that it is less than 21.9 inches long?
If one item is chosen at random, what is the probability that it is less than 21.9 inches long?
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Answered by
GPT-4o mini
To find the probability that a randomly selected item has a length less than 21.9 inches, we can use the properties of the normal distribution. We will first standardize the value using the Z-score formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
where:
- \( X \) is the value we are interested in (21.9 inches),
- \( \mu \) is the mean (16.9 inches),
- \( \sigma \) is the standard deviation (3.7 inches).
Substituting the values:
\[
Z = \frac{21.9 - 16.9}{3.7} = \frac{5.0}{3.7} \approx 1.3514
\]
Now that we have the Z-score, we can use the Z-table (or a calculator) to find the probability that corresponds to \( Z = 1.3514 \).
Looking at the Z-table or using a calculator, we find:
\[
P(Z < 1.3514) \approx 0.9115
\]
Thus, the probability that a randomly selected item is less than 21.9 inches long is approximately **0.9115** or **91.15%**.
So, the final answer is:
\[
\text{Probability} \approx 0.9115
\]
\[
Z = \frac{X - \mu}{\sigma}
\]
where:
- \( X \) is the value we are interested in (21.9 inches),
- \( \mu \) is the mean (16.9 inches),
- \( \sigma \) is the standard deviation (3.7 inches).
Substituting the values:
\[
Z = \frac{21.9 - 16.9}{3.7} = \frac{5.0}{3.7} \approx 1.3514
\]
Now that we have the Z-score, we can use the Z-table (or a calculator) to find the probability that corresponds to \( Z = 1.3514 \).
Looking at the Z-table or using a calculator, we find:
\[
P(Z < 1.3514) \approx 0.9115
\]
Thus, the probability that a randomly selected item is less than 21.9 inches long is approximately **0.9115** or **91.15%**.
So, the final answer is:
\[
\text{Probability} \approx 0.9115
\]
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