To find the probability that a randomly-selected adult male will have a height between 69.3 inches and 71.2 inches, we can standardize the values using the normal distribution. Given that the mean height (μ) is 69 inches and the standard deviation (σ) is 2.5 inches, we can use the z-score formula:
\[ z = \frac{(X - \mu)}{\sigma} \]
where \(X\) is the value we are standardizing.
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Calculate the z-scores for 69.3 inches and 71.2 inches:
For 69.3 inches: \[ z_1 = \frac{(69.3 - 69)}{2.5} = \frac{0.3}{2.5} = 0.12 \]
For 71.2 inches: \[ z_2 = \frac{(71.2 - 69)}{2.5} = \frac{2.2}{2.5} = 0.88 \]
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Look up the z-scores in the standard normal distribution table (or use a calculator):
- For \(z_1 = 0.12\), the cumulative probability \(P(Z < 0.12) \approx 0.5478\).
- For \(z_2 = 0.88\), the cumulative probability \(P(Z < 0.88) \approx 0.8121\).
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Calculate the probability of being between the two heights:
\[ P(69.3 < X < 71.2) = P(Z < 0.88) - P(Z < 0.12) \]
\[ P(69.3 < X < 71.2) = 0.8121 - 0.5478 = 0.2643 \]
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Convert this to a percentage:
\[ 0.2643 \times 100 \approx 26.43% \]
Based on the options given, the closest probability to approximately 26.43% is 26.28%.
So, the answer is: 26.28%.
2 answers