Respuesta :
Answer:
a) 0.194
b) $440.8
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = $9.40
Standard Deviation, σ = $4.40
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Standard error due to sampling =
[tex]\dfrac{\sigma}{\sqrt{n}} = \dfrac{4.40}{\sqrt{40}} = 0.6957[/tex]
a) P(tips from 40 parties > $400)
Thus, the average pay is
[tex]\dfrac{400}{40} = \$10[/tex]
P(x > 10)
[tex]P( x > 10) = P( z > \displaystyle\frac{10 - 9.40}{0.6957}) = P(z > 0.8624)[/tex]
[tex]= 1 - P(z \leq 0.8624)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 610) = 1 - 0.806 = 0.194 = 19.4\%[/tex]
b) We have to find the value of x such that the probability is 0.01
[tex]P( X > x) = P( z > \displaystyle\frac{x - 9.40}{0.6957})=0.01[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 9.40}{0.6957})=0.03 [/tex]
[tex]=P( z \leq \displaystyle\frac{x - 9.40}{0.6957})=0.99 [/tex]
Calculation the value from standard normal z table, we have,
[tex]\displaystyle\frac{x - 9,40}{0.6957} = 2.326\\\\x = 11.02[/tex]
Thus, the mean tip is $11.02
Tip for 40 parties =
[tex]11.02\times 40 = \$440.8[/tex]
Thus, he earns $440.8 on the best 1% of such weekends.
The probability that he will earn at least $400 is 19.4%. He earns $440.8 on the best 1% of such weekends.
What is normal a distribution?
It is also called the Gaussian Distribution. It is the most important continuous probability distribution. The curve looks like a bell, so it is also called a bell curve.
A waiter believes the distribution of his tips has a model that is slightly skewed to the left, with a mean of $9.40 and a standard deviation of $4.40.
He usually waits at about 40 parties over a weekend of work.
We know that the formula
[tex]z-score = \dfrac{x-\mu}{\sigma}[/tex]
Standard error due to sampling is given as
[tex]\rm Standard \ error = \dfrac{\sigma}{\sqrt{n}} = \dfrac{4.40}{\sqrt{40}}= 0.6957[/tex]
a) P(tips from 40 parties)
Thus, the average pay will be
[tex]\dfrac{400}{40} = \$10[/tex]
[tex]P(x > 10) = P(z > \dfrac{10-9.40}{0.6957}) = P(z > 0.864)\\\\P(x > 10) = 1- P(z\leq 0.8624)\\\\P(x > 10) = 0.194 = 19 .4\%[/tex]
b) We have to find the value of x such that the probability is 0.01
[tex]\rm P(X > x) = P(z > \dfrac{x-9.40}{0.9657}) = 0.01\\\\P(X > x) = 1 - P(z \leq \dfrac{x-9.40}{0.9657}) = 0.01\\\\P(X > x) = P(z \leq \dfrac{x-9.40}{0.9657}) = 0.99[/tex]
The value from the standard normal z-table, we have
[tex]\rm \dfrac{x-9.40}{0.6957} = 2.326\\\\x = 11.02[/tex]
Thus, the mean tip is $11.02
Tip for 40 parties = 11.02 × 40 = $440.8
He earns $440.8 on the best 1% of such weekends.
More about the normal distribution link is given below.
https://brainly.com/question/12421652
