Answer to Question #152008 in Statistics and Probability for JG

Question

1.    The lifetime of light bulbs produced by a company are normally distributed with mean 1000 hours and standard deviation 130 hours.

a)    The top 30% of all light bulbs should last at least how many hours?

b)    What is the probability that a randomly selected light bulb will last at least 900 hours?

c)    If we randomly select 8 light bulbs, what is the probability that the average of these light bulbs will last more than 1050 hours?

Answer 1

Let $X=$ the lifetime of light bulb in hours: $X\sim N(\mu, \sigma^2)$

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

Given $\mu=1000\ h, \sigma=130\ h$

a)

$$P(X\geq x_1)=0.3$$

$$P(X\geq x_1)=1-P(X<x_1)$$

$$=1-P(Z<\dfrac{x_1-1000}{130})=0.3$$

$$P(Z<\dfrac{x_1-1000}{130})=0.7$$

$$\dfrac{x_1-1000}{130}\approx0.524401$$

$$x_1=1000+130(0.524401)=1068$$

The top 30% of all light bulbs should last at least 1068 hours.

b)

$$P(X\geq 900)=1-P(X<900)$$

$$=1-P(Z<\dfrac{900-1000}{130})$$

$$\approx1-P(Z<-0.76923077)$$

$$\approx1-0.220878=0.779122$$

$0.779122$

c) Let $Y=$ the average lifetime of these bulbs in hours: $Y\sim N(\mu, \sigma^2/n)$

Then $Z=\dfrac{Y-\mu}{\sigma/\sqrt{n}}\sim N(0, 1)$

Given $\mu=1000\ h, \sigma=130\ h, n=10$

$$P(Y>1050)=1-P(Y\leq1050)$$

$$=1-P(Z\leq\dfrac{1050-1000}{130/\sqrt{10}})$$

$$\approx1-P(Z\leq1.21626064)$$

$$\approx1-0.888057=0.111943$$

$0.111943$