Question

A waiter believes the distribution of his tips has a model that is slightly skewed to the left​, with a mean of ​$8.90 and a standard deviation of ​$4.50. He usually waits on about 50 parties over a weekend of work. ​a) Estimate the probability that he will earn at least ​$550. ​b) How much does he earn on the best 10​% of such​ weekends?

Answer 1

Answer:

A. 0.3204    B. $14.669

Explanation:

Mean = 8.9      SD = 4.5

Required probability = P (X >/= 550/50)

P(X>/=11) = 1 - P[(X - mean/SD) < (11 - mean)/SD]

              = 1 - P(Z < (11-8.9)/4.5)

P(X>/=11) = 1 - P(Z < 0.4666667)

Using Excel NORMDIST(0.4666667,0,1,1)

P(X>/=11) = 1 - 0.6796 = 0.3204

The probability that she will earn at least $550 = 0.3204

b. P ( X  >  x )  =  0.10

1  −  P ( X  −  mean)/SD  ≤  (x  −  mean) /SD = 0.10

P ( Z  ≤  z )  =  0.90

Where,

z  =  (x  −  mean )/SD

Excel function for the value of z:

=NORMSINV(0.9)

=1.282

Hence (x - mean)/SD = 1.282

= (x - 8.9)/4.5 = 1.282

x = (1.282*4.5) + 8.9

x = 14.669

He earns $14.669 on the best 10% of such weekends.