Question

A particular fruit's weights are normally distributed, with a mean of 250 grams and a standard deviation of 19 grams.

If you pick one fruit at random, what is the probability that it will weigh between 262 grams and 281 grams

Answer 1

To calculate the probability that a randomly selected fruit will weigh between 262 grams and 281 grams, we can use the properties of the normal distribution.

Given the following parameters:
- Mean (\(\mu\)): 250 grams
- Standard deviation (\(\sigma\)): 19 grams

To find this probability, we will:
1. Convert the weights into z-scores using the formula:

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

2. Use the z-scores to find the corresponding cumulative probabilities from the standard normal distribution table.

**Step 1: Calculate z-scores**

For \(X = 262\) grams:

\[
z_1 = \frac{262 - 250}{19} = \frac{12}{19} \approx 0.6316
\]

For \(X = 281\) grams:

\[
z_2 = \frac{281 - 250}{19} = \frac{31}{19} \approx 1.6316
\]

**Step 2: Find cumulative probabilities**

Using a standard normal distribution table (or a calculator), we find:

- The cumulative probability for \(z_1\) (approximately 0.63):
- \(P(Z < 0.63) \approx 0.7357\)

- The cumulative probability for \(z_2\) (approximately 1.63):
- \(P(Z < 1.63) \approx 0.9484\)

**Step 3: Calculate the probability between the two z-scores**

To find the probability that the weight of the fruit is between 262 grams and 281 grams, we subtract the cumulative probabilities:

\[
P(262 < X < 281) = P(Z < 1.6316) - P(Z < 0.6316) \approx 0.9484 - 0.7357
\]

\[
P(262 < X < 281) \approx 0.2127
\]

Thus, the probability that a randomly selected fruit will weigh between 262 grams and 281 grams is approximately **0.2127**, or **21.27%**.