To find the probability that a randomly selected plant is between 20.2 cm and 24.9 cm tall, we first need to calculate the z-scores for both of these heights using the mean and standard deviation given.
-
Calculate the z-scores:
The formula for calculating the z-score is:
\[ z = \frac{(X - \mu)}{\sigma} \]
where:
- \(X\) is the value we are evaluating,
- \(\mu\) is the mean,
- \(\sigma\) is the standard deviation.
For \(X = 20.2\) cm: \[ z = \frac{(20.2 - 20)}{3} = \frac{0.2}{3} \approx 0.0667 \]
For \(X = 24.9\) cm: \[ z = \frac{(24.9 - 20)}{3} = \frac{4.9}{3} \approx 1.6333 \]
-
Find the probabilities corresponding to these z-scores using the z-score table:
-
For \(z \approx 0.0667\), looking up \(z = 0.06\) and \(z = 0.07\):
- \(P(Z < 0.06) \approx 0.5239\)
- \(P(Z < 0.07) \approx 0.5279\)
To estimate for \(z \approx 0.0667\), we can interpolate: \[ P(Z < 0.0667) \approx 0.5239 + \frac{(0.0667 - 0.06)}{(0.07 - 0.06)} \times (0.5279 - 0.5239) \] \[ \approx 0.5239 + 0.67 \times 0.004 = 0.5239 + 0.00268 \approx 0.52658 \]
-
For \(z \approx 1.6333\), looking up \(z = 1.63\) and \(z = 1.64\):
- \(P(Z < 1.63) \approx 0.9484\)
- \(P(Z < 1.64) \approx 0.9495\)
To estimate for \(z \approx 1.6333\), we can interpolate: \[ P(Z < 1.6333) \approx 0.9484 + \frac{(1.6333 - 1.63)}{(1.64 - 1.63)} \times (0.9495 - 0.9484) \] \[ \approx 0.9484 + 0.3 \times 0.0011 = 0.9484 + 0.00033 \approx 0.94873 \]
-
-
Calculate the probability of being between the two z-scores:
Now we find the probability that a plant is between 20.2 cm and 24.9 cm tall: \[ P(20.2 < X < 24.9) = P(Z < 1.6333) - P(Z < 0.0667) \approx 0.94873 - 0.52658 \approx 0.42215 \]
-
Convert to percentage:
To convert to a percentage: \[ 0.42215 \times 100 \approx 42.215 \]
Rounding to the hundredths place gives: \[ 42.22% \]
Thus, the probability that a randomly selected plant is between 20.2 cm and 24.9 cm tall is 42.22%.