Question

Suppose you are interested in how long it takes to get your food at a restaurant. Now, suppose this distribution is approximately normal with an average of eight minutes and a standard deviation of two minutes. If you made a control chart for this data, what would be the highest control limit? Using the above scenario, suppose someone gets the food after exactly eleven minutes. How many standard deviations from the mean is this value of eleven? Using the above scenario, what is the probability that someone would get the food after more than eleven minutes?

[I know the answer involves P(X>11) = P(X>1.5) = 0.5-0.4332 = 0.0668, what I don't understand is how you get the values of P(X>11) or P(X>1.5) or where 0.4332 comes from]

Answer 1

Answer: The probability that someone would get the food after more than eleven minutes is 0.0668

Given

u=mean	8
sd=standard deviation	2

We have to calculate the probability that someone would get the food after more than eleven minutes that is P(X>11)

x	11
Using(x-u)/sd=(11-8)/2	1.5
Using =normsdist(1.5)	0.933193
step 1-0.933193	0.066807

......................................................................................................................................

I know the answer involves P(X>11) = P(X>1.5) = 0.5-0.4332 = 0.0668, what I don't understand is how you get the values of P(X>11) or P(X>1.5) or where 0.4332 comes from]

The values 0.0668 is comes from the standard normal table. Open the stanadard normal table and check the value at 1.5.

I will suggest using excel command "=Normsdist(1.5)". This command will give the same value which you find in the table.

Paste the command in bold in excel and you will get your answer.

Hope this helps you in understanding otherwise do the comment. Thank You!