91Ó°ÊÓ

Probability: Probability means possibility. It is a branch of mathematics which deals with the occurrence of a random event.

(a)

Considering the given information:

\(\begin{aligned}n &= 500\\ p &= 75\% \\ &= 0{\rm{.75}}\end{aligned}\)

The following are the requirements for a normal binomial distribution approximation:

As a result, the requirements are met, and we can use the normal distribution to approximate the binomial distribution.

The z-score is the value (using the continuity correction) multiplied by the standard deviation\(\sqrt {{\rm{npq}}} {\rm{ = }}\sqrt {{\rm{np(1 - p)}}} \)and divided by the mean np.

\(\begin{aligned}z &= \frac{{x - np}}{{\sqrt {np(1 - p)} }}\\ &= \frac{{359.5 - 500(0.75)}}{{\sqrt {500(0.75)(1 - 0.75)} }} \approx - 1.60\\z &= \frac{{x - np}}{{\sqrt {np(1 - p)} }}\\ &= \frac{{400.5 - 500(0.75)}}{{\sqrt {500(0.75)(1 - 0.75)} }} \approx 2.63\end{aligned}\)

Using table A.3, calculate the corresponding normal probability:

\(\begin{aligned}P(360 \le X \le 400) &= P(359.5 < X < 400.5)\\ &= P( - 1.60 < Z < 2.63)\\ &= P(Z < 2.63) - P(Z < - 1.60)\\ &= 0.9957 - 0.0548\\ &= 0.9409\end{aligned}\)

Therefore, the required value is \({\rm{0}}{\rm{.9409}}\).

(b)

Considering the given information:

The z-score is the value (using the continuity correction) multiplied by the standard deviation \(\sqrt {{\rm{npq}}} {\rm{ = }}\sqrt {{\rm{np(1 - p)}}} \) and divided by the mean np.

\(\begin{aligned}z &= \frac{{x - np}}{{\sqrt {np(1 - p)} }}\\ &= \frac{{399.5 - 500(0.75)}}{{\sqrt {500(0.75)(1 - 0.75)} }} \approx 2.53\end{aligned}\)

Using table A.3, calculate the corresponding normal probability:

\(\begin{aligned}P(X < 400) &= P(X < 399 {\rm{.5)}}\\ &= P(Z < 2 {\rm{.53)}}\\ &= 0 {\rm{.9943}}\end{aligned}\)

Therefore, the required value is \({\rm{0}}{\rm{.9943}}\).