Answer to Question #255306 in Statistics and Probability for Bronkolly

Question

The time taken to complete a web-based questionnaire is normally distributed, with an average time (μ) of 9 minutes and a standard deviation (σ) of 1.55 minutes. What is the probability that a randomly selected person will take:

Answer 1

Let $X=$ the time taken to complete a web-based questionnaire: $X\sim N(\mu, \sigma^2).$

Given $\mu=9\ min, \sigma=1.55\ min.$

$$P(8<X<8.5)=P(X<8.5)-P(X\leq 8)$$

$$=P(Z<\dfrac{8.5-9}{1.55})-P(Z\leq \dfrac{8-9}{1.55})$$

$$\approx P(Z<-0.32258)-P(Z\leq -0.64516)$$

$$\approx0.3735064-0.2594113=0.114095$$

The probability that a randomly selected person will take between 8 and 8.5 minutes to complete the questionnaire is $0.114095.$

$$P(8.75<X<9.75)=P(X<9.75)-P(X\leq 8.75)$$

$$=P(Z<\dfrac{9.75-9}{1.55})-P(Z\leq \dfrac{8.75-9}{1.55})$$

$$\approx P(Z<0.48387)-P(Z\leq -0.16129)$$

$$\approx0.6857613-0.4359325=0.249829$$

The probability that a randomly selected person will take between 8.75 and 9.75 minutes to complete the questionnaire is $0.249829.$