Question

The mean height of a group of plants is 20 centimeters, with a standard deviation of 3 centimeters. What is the probability that a randomly selected plant is between 20.2 cm and 24.9 cm tall? Give the answer as a percentage to the hundredths place. (1 point) The probability that a randomly selected plant is between 20.2 and 24.9 centimeters tall is %.

Answer 1

To find the probability that a randomly selected plant is between 20.2 cm and 24.9 cm tall, we can use the properties of the normal distribution, based on the information given.

1. **Calculate the Z-scores** for the heights of 20.2 cm and 24.9 cm.

The formula for calculating the Z-score is:

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

where:
- \(X\) is the value for which we are finding the Z-score,
- \(\mu\) is the mean,
- \(\sigma\) is the standard deviation.

Given:
- \(\mu = 20\) cm,
- \(\sigma = 3\) cm.

For \(X = 20.2\) cm:

\[
Z_{20.2} = \frac{(20.2 - 20)}{3} = \frac{0.2}{3} \approx 0.0667
\]

For \(X = 24.9\) cm:

\[
Z_{24.9} = \frac{(24.9 - 20)}{3} = \frac{4.9}{3} \approx 1.6333
\]

2. **Use a Z-table** or a calculator to find the probabilities corresponding to these Z-scores.

- For \(Z_{20.2} \approx 0.0667\), the corresponding probability (from the Z-table or calculator) is approximately \(0.5279\).
- For \(Z_{24.9} \approx 1.6333\), the corresponding probability is approximately \(0.9484\).

3. **Calculate the probability** that a randomly selected plant is between these two heights:

\[
P(20.2 < X < 24.9) = P(Z < 1.6333) - P(Z < 0.0667)
\]

\[
P(20.2 < X < 24.9) \approx 0.9484 - 0.5279 = 0.4205
\]

4. **Convert the probability to a percentage**:

\[
0.4205 \times 100 \approx 42.05\%
\]

So, the probability that a randomly selected plant is between 20.2 cm and 24.9 cm tall is **42.05%**.