Question

Assume students spend an average of $20.00 on a meal at a restaurant in 2018. Assume the amount of money spent on a restaurant meal is normally distributed and that the standard deviation is $4.

What is the probability a randomly selected student spent more than $22?

What is the probability that a randomly selected student spent between $19 and $21?

Between what two values will the middle 99% of the amounts of money spent fall?

Answer 1

for normal distribution z score =(X-μ)/σx

1) probability a randomly selected student spent more than $22:

probability =

P(X>22)

=

P(Z>0.5)=

1-P(Z<0.5)=

1-0.6915=

0.3085

2)

probability that a randomly selected student spent between $19 and $21 :

probability =

P(19<X<21)

=

P(-0.25<Z<0.25)=

0.5987-0.4013=

0.1974

3)

for middle 99% values critical z =2.576

therefore middle 99% of the amounts of money spent fall between 20-/+2.576*4 = 9.68 and 30.32