Answer to Question #249177 in Statistics and Probability for Bongs

Question #249177

A pizza parlour will deliver a pizza take-away order to the customer if s/he lives within a 5 kilometre radius from the pizza parlour. If the customer lives within this radius, it is found that the time taken to receive the pizza is normally distributed with a mean time of 45 minutes and a standard deviation of 8 minutes. Paul has relocated and now lives within the 5 kilometre radius. He orders a pizza from the pizza parlour. The probability that it takes between 43 and 49 minutes for him to receive his pizza is


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Expert's answer
2021-10-12T14:09:03-0400

Let X=X= the time taken to receive the pizza: XN(μ,σ2).X\sim N(\mu, \sigma^2).

Given μ=45 min,σ=8 min.\mu=45\ min, \sigma=8\ min.

P(43<X<49)=P(X<49)P(X43)P(43<X<49)=P(X<49)-P(X\leq 43)

=P(Z<49μσ)P(Z43μσ)=P(Z< \dfrac{49-\mu}{\sigma})-P(Z\leq \dfrac{43-\mu}{\sigma})

=P(Z<49458)P(Z43458)=P(Z< \dfrac{49-45}{8})-P(Z\leq \dfrac{43-45}{8})

=P(Z<0.5)P(Z0.25)=P(Z< 0.5)-P(Z\leq-0.25)

0.691462460.40129367\approx0.69146246-0.40129367

0.290169\approx0.290169


The probability that it takes between 43 and 49 minutes for him to receive his pizza is 0.290169.0.290169.


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