Question

Find the probability that a randomly chosen score from a normal distribution with a mean of 85 and a standard deviation of 12 is between 78 and 92.

Answer 1

To find the probability that a randomly chosen score from a normal distribution with a mean of 85 and a standard deviation of 12 falls between 78 and 92, we first need to standardize the scores (convert them to z-scores).

The z-score is calculated using the formula:
\[
z = \frac{(X - \mu)}{\sigma}
\]
where:
- \(X\) is the value from the distribution,
- \(\mu\) is the mean,
- \(\sigma\) is the standard deviation.

1. **Calculate the z-scores for 78 and 92**:

- For \(X = 78\):
\[
z_{78} = \frac{(78 - 85)}{12} = \frac{-7}{12} \approx -0.5833
\]

- For \(X = 92\):
\[
z_{92} = \frac{(92 - 85)}{12} = \frac{7}{12} \approx 0.5833
\]

2. **Use the z-table to find the probabilities**:

Now, we will look up the z-scores in the standard normal distribution table (z-table) to find the probabilities corresponding to these z-scores.

- For \(z \approx -0.5833\):
The approximate table value (or using a calculator) gives:
\[
P(Z < -0.5833) \approx 0.2794
\]

- For \(z \approx 0.5833\):
The approximate table value gives:
\[
P(Z < 0.5833) \approx 0.7206
\]

3. **Calculate the probability that the score is between 78 and 92**:

To find the probability that \(X\) is between 78 and 92, we take the difference of the two probabilities:
\[
P(78 < X < 92) = P(Z < 0.5833) - P(Z < -0.5833) \approx 0.7206 - 0.2794 = 0.4412
\]

Therefore, the probability that a randomly chosen score from this normal distribution is between 78 and 92 is approximately **0.4412** or **44.12%**.