Answer to Question #107230 in Statistics and Probability for sadia

Question

The weights of bananas in a fruit shop have a normal distribution with mean 150 grams
and standard deviation 50 grams. Three sizes of bananas are sold.
Small: under 95 grams Medium: between 95 grams and 205 grams Large: over 205 grams
(i) Find the proportion of bananas that are small. [4]
(ii) Find the weight exceeded by 10% of the bananas. [4]
The price of bananas are 10 cents, 20 cents for a medium banana and 25 cents for a
large banana.
(iii) Show that a randomly chosen banana costs 20 cents is 0.7286. [4]

Answer 1

$i) F(x)=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{x}e^{-\frac{(t-a)^2}{2\sigma^2}}dt.\\ a=150\\ \sigma=50\\ F(x)=\frac{1}{\sqrt{2\pi}50}\int_{-\infty}^{x}e^{-\frac{(t-150)^2}{2\cdot 50^2}}dt.$

So we should find $F(95).$

$F(95)=\frac{1}{\sqrt{2\pi}50}\int_{-\infty}^{95}e^{-\frac{(t-150)^2}{2\cdot 50^2}}dt=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{-1.1}e^{-\frac{z^2}{2}}dz\approx\frac{1}{2}-0.3643=0.1357$ (here we used substitution $z=\frac{t-150}{50}$ ).

$ii) P\{\xi>w\}=0.1$

We should find $w$.

$P\{\xi>w\}=1-F(w)=0.1\\ F(w)=0.9.\\ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\frac{w-150}{50}}e^{-\frac{z^2}{2}}dz=0.9\\ 0.5+\frac{1}{\sqrt{2\pi}}\int_{0}^{\frac{w-150}{50}}e^{-\frac{z^2}{2}}dz=0.9\\ \frac{1}{\sqrt{2\pi}}\int_{0}^{\frac{w-150}{50}}e^{-\frac{z^2}{2}}dz=0.4\\ \frac{w-150}{50}=1.28\\ w=214.$

$iii)$ We should find the proportion of bananas that are medium. The proportion equals

$F(205)-F(95)\approx \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{1.1}e^{-\frac{z^2}{2}}dz-0.1357\approx 0.5+0.3643-0.1357=0.7286.$