The weights of the students of a college are normally distributed with standard deviation 5 kg. If 10% of the students weigh more than 72 kg, then the mean weight of the students is (Given \(P(0 \le Z \le 1.28) = 0.40\), where Z is the standard normal variate)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 65.6 kg
Let X be the weight and \(\mu\) the mean. The area to the right of the mean is 0.5, and 10% = 0.10 lies beyond 72 kg, so the area between the mean and 72 kg is
\(0.5 - 0.10 = 0.40\)
From the table, the area 0.40 corresponds to Z = 1.28, so 72 kg is 1.28 standard deviations above the mean:
\( \displaystyle \frac{72 - \mu}{5} = 1.28 \)
\(72 - \mu = 6.4\)
\(\mu = 65.6\)
Why the other options are incorrect
Option (1) 78.4 kg — Incorrect. This is 72 + 6.4; a mean above 72 kg would put more than half the students above 72 kg, not 10%.
Option (2) 67 kg — Incorrect. This takes 72 kg to be exactly one standard deviation (5 kg) above the mean; it is 1.28 standard deviations above.
Option (4) 70 kg — Incorrect. With mean 70 kg, Z for 72 kg is 0.4, and far more than 10% would weigh more than 72 kg.