A random variable is normally distributed with a mean of 0 and a standard deviation of 1 - HSC - SSCE Mathematics Standard - Question 38 - 2023 - Paper 1

From the table, the probability corresponding to ( z = 1.33 ) is 0.9082.

Thus, the probability of a koala weighing less than 11.93 kg is ( P(X < 11.93) = 0.9082 ).

To find the probability of a koala weighing more than 11.93 kg, use:

$P(X > 11.93) = 1 - P(X < 11.93)$

Therefore:

$P(X > 11.93) = 1 - 0.9082 = 0.0918.$

In a group of 400 adult male koalas, the expected number that weighs more than 11.93 kg is:

$\text{Number of koalas} = P(X > 11.93) \times 400 = 0.0918 \times 400 = 36.72.$

Rounding down, we expect approximately 36 koalas.