Questions: A manufacturer knows that their items have a normally distributed length, with a mean of 17.3 inches, and standard deviation of 5.5 inches. If one item is chosen at random, what is the probability that it is less than 20.9 inches long?

A manufacturer knows that their items have a normally distributed length, with a mean of 17.3 inches, and standard deviation of 5.5 inches.

If one item is chosen at random, what is the probability that it is less than 20.9 inches long?

Transcript text: A manufacturer knows that their items have a normally distributed length, with a mean of 17.3 inches, and standard deviation of 5.5 inches. If one item is chosen at random, what is the probability that it is less than 20.9 inches long? $\square$ Submit Question

Solution

Solution Steps

Step 1: Define the Normal Distribution Parameters

The length of the items produced by the manufacturer follows a normal distribution characterized by the mean $ \mu = 17.3 $ inches and the standard deviation $ \sigma = 5.5 $ inches.

Step 2: Calculate the Z-Score

To find the probability that a randomly chosen item is less than $ 20.9 $ inches long, we first calculate the Z-score for $ 20.9 $ inches using the formula:

\[ Z = \frac{X - \mu}{\sigma} \]

Substituting the values:

\[ Z_{end} = \frac{20.9 - 17.3}{5.5} = \frac{3.6}{5.5} \approx 0.6545 \]

Step 3: Determine the Probability

The probability that a randomly chosen item is less than $ 20.9 $ inches long can be expressed using the cumulative distribution function $ \Phi $:

\[ P(X < 20.9) = \Phi(Z_{end}) - \Phi(Z_{start}) \]

Since $ Z_{start} $ corresponds to negative infinity, we have:

\[ P(X < 20.9) = \Phi(0.6545) - \Phi(-\infty) = \Phi(0.6545) - 0 \]

From the calculations, we find:

\[ P(X < 20.9) \approx 0.7436 \]