Answer to Question #89610 in Statistics and Probability for Johnathan

Question

5.2 The service time of the first service of a BMW is found to be normally distributed, with a mean of 70 minutes and a standard deviation of 9 minutes. Determine the following:
5.2.1 The probability that the first service will take more than an hour and a half.
5.2.2 The probability that the first service will take between 50 and 60 minutes.
i need urgent help. thank you

Answer 1

If $X\sim N(\mu, \sigma^2)$ then

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

Given that $\mu=70 \ min, \sigma=9 \ min.$

5.2.1 The probability that the first service will take more than an hour and a half. 

$$(1+1/2)\ h=90\ min$$

$$Z={90-70\over 9}={20\over 9}\approx 2.2222$$

$$P(X>90)=P(Z>{20\over 9})=1-P(Z\leq{20\over 9})=$$

$$=1-0.986869=0.013131$$

The probability that the first service will take more than an hour and a half is $0.013131.$

5.2.2 The probability that the first service will take between 50 and 60 minutes. 

$$Z_1={60-70\over 9}=-{10\over 9}\approx -1.1111$$

$$Z_2={50-70\over 9}=-{20\over 9}\approx -2.2222$$

$$P(50<X<60)=P({-20\over 9}<Z<{-10\over 9})=$$

$$=P(Z<-1.1111)-P(Z<-2.2222)\approx$$

$$\approx 0.133260-0.013134=0.120126$$

The probability that the first service will take between 50 and 60 minutes is $0.120126.$