Question #255306

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:

Expert's answer

Let X=X= the time taken to complete a web-based questionnaire: X∼N(μ,σ2).X\sim N(\mu, \sigma^2).

Given μ=9 min,σ=1.55 min.\mu=9\ min, \sigma=1.55\ min.


P(8<X<8.5)=P(X<8.5)−P(X≤8)P(8<X<8.5)=P(X<8.5)-P(X\leq 8)

=P(Z<8.5−91.55)−P(Z≤8−91.55)=P(Z<\dfrac{8.5-9}{1.55})-P(Z\leq \dfrac{8-9}{1.55})

≈P(Z<−0.32258)−P(Z≤−0.64516)\approx P(Z<-0.32258)-P(Z\leq -0.64516)

≈0.3735064−0.2594113=0.114095\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 0.114095.0.114095.




P(8.75<X<9.75)=P(X<9.75)−P(X≤8.75)P(8.75<X<9.75)=P(X<9.75)-P(X\leq 8.75)

=P(Z<9.75−91.55)−P(Z≤8.75−91.55)=P(Z<\dfrac{9.75-9}{1.55})-P(Z\leq \dfrac{8.75-9}{1.55})

≈P(Z<0.48387)−P(Z≤−0.16129)\approx P(Z<0.48387)-P(Z\leq -0.16129)

≈0.6857613−0.4359325=0.249829\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 0.249829.0.249829.



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