Answer to Question #107353 in Statistics and Probability for Keiron

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. [

Answer 1

Let $X=$ the weight of bananas in grams: $X\sim N(\mu, \sigma^2).$

Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0, 1)$

Given that $\mu=150\ g,\sigma=50\ g$

(i)

$$P(X<95)=P(Z<{95-150\over 50})=P(Z<-1.1)\approx0.135666$$

13.5666% of bananas are small

(ii)

$$P(X>X^*)=P(Z>Z^*)=0.1$$

$$P(Z\leq Z^*)=1-P(Z>Z^*)=1-0.1=0.9=>$$

$$=>Z^*\approx1.281552={X^*-150\over 50}=>X^*\approx214\ g$$

The weight exceeded by 10% of the bananas is 214 grams.

(iii) The price of medium bananas are 20 cents.

$$P(95<X<205)=P(X<205)-P(X<95)=$$

$$=P(Z<{205-150\over 50})-P(Z<{95-150\over 50})=$$

$$=P(Z<1.1)-P(Z<-1.1)\approx$$

$$\approx0.864334-0.135666\approx0.728668$$