Question

A study found that the mean amount of time cars spent in​ drive-throughs of a certain​ fast-food restaurant was 1.583 seconds. Assuming​ drive-through times are normally distributed with a standard deviation of ​30 seconds, complete parts​ (a) through​ (d) below.

​(a) The probability that a randomly selected car will get through the​ restaurant's drive-through in less than 106 seconds is________

​(Round to four decimal places as​ needed.)

(b)The probability that a randomly selected car will spend more than 215 seconds in the​ restaurant's drive-through is______

​​(Round to four decimal places as​ needed.)

(c) The proportion of cars that spend between 2 and 3 minutes in the​ restaurant's drive-through is______

​​(Round to four decimal places as​ needed.)

(d) The probability that a car spends more than 3 minutes in the​ restaurant's drive-through is_____so is______be​ unusual, since the probability is_______than 0.05.

​​(Round to four decimal places as​ needed.)

Answer 1

Solution:

Given:

First need to conver the population mean from second to minutes.

Population mean= 1.583 seconds = 1.583*60 = 94.98 seconds

Therefore, μ=94.98 and σ=30

a) We have to find P(X<106) = ...?

P(Z<0.3673)=0.6433 ...Using excel formula, =NORMSDIST(0.3673)

Hence, P(X<106) = 0.6433

b) We have to find P(X<215) = ...?

P(Z>4.0007)=1-P(Z<4.0007)

P(Z>4.0007) =1-1 = 0 ...Using excel formula, =NORMSDIST(4.007)

Hence, P(X<215) = 0

c) Here first need to convert given standard deviation into minutes.

σ=30/60 = 0.5

Therefore, μ=1.583 and σ=0.5

We have to find P(2X3) = ...?

P(0.834≤Z≤2.834) = P(Z≤2.834)−P(Z≤0.834)

P(0.834≤Z≤2.834) = 0.9977−0.7979=0.1998

...Using excel formula, =NORMSDIST(2.834) and =NORMSDIST(0.834)

Hence, P(2X3) = 0.1998

d)

P(X>3) = ...?

P(Z>2.834)=1-P(Z<2.834)

P(Z>2.834)=1−0.9977=0.0023 ...Using excel formula, =NORMSDIST(2.834)

Hence, P(X>3) = 0.0023

The probability that a car spends more than 3 minutes in the​ restaurant's drive-through is 0.0023 so is should be unusual, since the probability is less than 0.05.

Done