5.4.4 Probability Distributions, Long Questions

Question 10:
(a) It is found that 60% of the students from a certain class obtained grade A in English in O level trial examination.
If 10 students from the class are selected at random, find the probability that
(i) exactly 7 students obtained grade A.
(ii) not more than 7 students obtained grade A.

(b) Diagram below shows a standard normal distribution graph representing the volume of soy sauce in bottles produced by a factory.

It is given the mean is 950 cm³ and the variance is 256 cm⁶. If the percentage of the volume more than V is 30.5%, find
(i) the value of V,
(ii) the probability that the volume between 930 cm³ and 960 cm³.

Solution:
$\begin{array}{l}\text{(a)(i)}\\ P(X=r)={}^{n}c{}_r.p^r.q^{n-r}\\ P(X=7)={}^{10}C{}_7{\left(0.6\right)}^7{\left(0.4\right)}^3\\ =0.0860\\ \\ \left(ii\right)\\ P(X\le 7)\\ =1-P(X>7)\\ =1-P\left(X=8\right)-P\left(X=9\right)-P\left(X=10\right)\\ =1-{}^{10}C{}_8{\left(0.6\right)}^8{\left(0.4\right)}^2-{}^{10}C{}_9{\left(0.6\right)}^9{\left(0.4\right)}^1-{}^{10}C{}_{10}{\left(0.6\right)}^{10}{\left(0.4\right)}^0\\ =1-0.1209-0.0403-0.0060\\ =0.8328\end{array}$


$\begin{array}{l}\text{(b)}\left(i\right)\\ P\left(X>V\right)=30.5\%\\ P\left(Z>\frac{V-950}{16}\right)=0.305\\ P\left(Z>0.51\right)=0.305\\ \frac{V-950}{16}=0.51\\ V=0.51\left(16\right)+950\\ =958.16{\text{ cm}}^3\end{array}$

$\begin{array}{l}\left(ii\right)\\ \text{Probability}\\ =P\left(930<X<960\right)\\ =P\left(\frac{930-950}{16}<Z<\frac{960-950}{16}\right)\\ =P\left(-1.25<Z<0.625\right)\\ =1-P\left(Z>1.25\right)-P\left(Z>0.625\right)\\ =1-0.1056-0.2660\\ =0.6284\end{array}$