Question

The lengths of rods from a certain factory are normally distributed with mean xcm and standard deviation 6cm. It is known that 4.78% of the rods have lengths greater than 82cm. Find to the nearest integer,
(i) the mean x.
(ii) the range, s (symmetrical about the mean) within which 75% of the lengths of the rods lie.
(iii) the probability that a rod chosen at random has a length between 62cm and 72cm.

Answer 1

(i) Let $Y=$ the length of rod: $Y\sim N(x, \sigma^2).$ Then $Z=\dfrac{Y-x}{\sigma}\sim N(0, 1)$

Given $\sigma=6\ cm.$

$$P(Y>82)=1-P(Y\leq82)$$

$$=1-P(Z\leq\dfrac{82-x}{6})=0.0478$$

$$P(Z\leq\dfrac{82-x}{6})=0.9522$$

$$\dfrac{82-x}{6}\approx1.66657$$

$$x\approx72$$

$mean=x=72$

(ii)

$$P(x-\dfrac{s}{2}<Y<x+\dfrac{s}{2})$$

$$=P(Y<x+\dfrac{s}{2})-P(Y\leq x-\dfrac{s}{2})=$$

$$=P(Z<\dfrac{\dfrac{s}{2}}{6})-P(Z\leq\dfrac{-\dfrac{s}{2}}{6})=0.75$$

$$P(Z\leq-\dfrac{s}{12})=\dfrac{1-0.75}{2}$$

$$-\dfrac{s}{12}\approx-1.15035$$

$$s\approx13.0842$$

(iii)

$$P(62<Y<72)=P(Y<72)-P(Y\leq62)$$

$$=P(Z<\dfrac{72-72}{6})-P(Z\leq\dfrac{62-72}{6})$$

$$=P(Z<0)-P(Z\leq-\dfrac{5}{3})\approx0.5-0.04779$$

$$=0.45221$$

The probability that a rod chosen at random has a length between 62cm and 72cm is 0.45221.