Question

A waiter estimates that his average tip per table is $18 with a standard deviation of $4. If his tables seat 8 customers, and the tip is normally distributed: - What is the probability that the average tip for one table is less than $20? - What is the probability that the average tip is more than $20? - What is the probability that the tip is equal to 21? - What is the probability that the average tip for one table is between $18 and $20?

Answer 1

µ = 18, σ = 4, n = 8

a) P(X̅ < 20) =

= P( (X̅-μ)/(σ/√n) < (20-18)/(4/√8) )

= P(z < 1.4142)

Using excel function:

= NORM.S.DIST(1.4142, 1)

= 0.9214

b) P(X̅ > 20) =

= P( (X̅-μ)/(σ/√n) > (20-18)/(4/√8) )

= P(z > 1.4142)

= 1 - P(z < 1.4142)

Using excel function:

= 1 - NORM.S.DIST(1.4142, 1)

= 0.0786

c)

P(18 < X̅ < 20) =

= P( (18-18)/(4/√8) < (X-µ)/(σ/√n) < (20-18)/(4/√8) )

= P(0 < z < 1.4142)

= P(z < 1.4142) - P(z < 0)

Using excel function:

= NORM.S.DIST(1.4142, 1) - NORM.S.DIST(0, 1)

= 0.4214