Question

A waitress believes the distribution of her tips has a model that is slightly skewed to the left​, with a mean of ​$9.20 and a standard deviation of ​$5.80. She usually waits on about 30 parties over a weekend of work.
​A) Estimate the probability that she will earn at least ​$350.
​B) How much does she earn on the best 10​% of such​ weekends?

Answer 1

Solutions:-

(a):-

The sample of 30 parties as larg enough. the mean trip is with standard deviation since the conditions are satisfied the CLT allow us to model the sampling distribution y- bar which a normal model with y = $9.20

and standard deviation y = 5.80 / √30 = $1.0590

In order to earn atleast $350 the batrees would have to average 350 / 30 = 11.67 per party.

Now,

since and we have.

P(X > 11.67) = P (X - > 11.67 - 9.20 )

= P{( X - ) / > ( 11.67 - 9.20) / 1.0590 )

since Z = {(X - ) / } and (11.67 - 9.20) / 1.0590

= 2.3324

P(X >11.67) = P ( Z > 2.3324)

use the standard normal table to conclude that :

(b):-

The best 10% of weekends indicate that the probability or area under the normal is 0.90.

Z{probability(Area)} = 1.645,

Use the excel formula *=NORMSINV(0.90)*

Thus,

z = (x-bar -) /

1.645 =( x - 9.20) / 1.0590

x-bar = 9.20 + 1.645 × 1.0590

= 10.9421

Mean trip = $10.9421

Tip for 30 parties = 10.9421 × 30

= $328.3 in the best 1% of soch weekend.

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