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One year consumers spent an average of $23 on a meal at a restaurant. Assume that...

One year consumers spent an average of $23 on a meal at a restaurant. Assume that the amount spent on a restaurant meal is normally distributed and that the standard deviation is $6.

What is the probability that a randomly selected person spent more than $28?

P(X>$28)=?

​

Math   Statistics-And-Probability   
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Answer #1

given data are

population mean =$23

population sd =$6

sample n=1

sample mean =$28

in a z normal distribution, z=

z=(28-23)/(6/)

= 5/6

=0.83333= 0.83

p (X>28)=p (z >0.83)

= 1- p (z0.83)

= 1-0.7967

=0.2033

p (X>28)= 0.2033

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