Question

Assume that the weights of 18-year-old males in a population follow a normal distribution with mean 65 kg and standard deviation 11 kg.Eight boys are chosen at random from the population. Find the probability that at most three of them weigh at least 70 kg.

Answer 1

The Solution:

Using the formula:

$Z_{70}=\frac{x-\mu}{\sigma}$

Where,

$\begin{gathered} x=70kg \\ \mu=65kg \\ \sigma=11kg \end{gathered}$

Substituting, we get

$P(Z>Z_{70})=\frac{70-65}{11}=\frac{5}{11}=0.45455$

From the Z score table,

$\begin{gathered} P(Z>Z_{70})=0.3242=p \\ 1-P(Z>Z_{70})=1-0.3242=0.6758=q \end{gathered}$

By the Binomial expansion formula,

$(p+q)^8=+\cdots+^8C_3p^3\cdot q^5+^8C_2p^2\cdot q^6+^8C_1p^1\cdot q^7+^8C_0p^0\cdot q^8$

So, the probability that at most three of them weigh at least 70 kg is

$\begin{gathered} ^8C_3(0.3242)^3\cdot(0.6758)^5+^8C_2(0.3242)^2\cdot(0.6758)^6+ \\ ^8C_1(0.3242)^1\cdot(0.6758)^7+^8C^{}_0(0.6758)^8 \end{gathered}$$=0.49562\approx0.4956$

Therefore, the correct answer is 0.4956