"A waiter believes the distribution of his tips has a model that is slightly skewed to the left​, with a mean of ​$10.60 and a standard deviation of ​$6.60. He usually waits on about 50 parties over a weekend of work. ​a) Estimate the probability that he will earn at least ​$600. ​b) How much does he earn on the best 1​% of such​ weekends?"

Answers

Answer 1

Answer:

(a) 0.0668

(b) $638.74

Step-by-step explanation:

Let X denote the tips earned by a waiter.

It is provided that X follows a left-skewed distribution with mean, μ = $10.60 and standard deviation, σ = $6.60.

It is also provided that, the waiter usually waits on about n = 50 parties over a weekend of work.

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

 [tex]\mu_{S}=n\mu\\[/tex]

And the standard deviation of the distribution of the sum of values of X is given by,  

[tex]\sigma_{S}=\sqrt{n}\sigma[/tex]

As the sample size is large, i.e. n = 50 > 30, the Central Limit Theorem can be used to approximate the sampling distribution of total tips by the normal distribution.

The mean and standard deviation are:

[tex]\mu_{S}=50\times 10.60=530\\\\\sigma_{S}=\sqrt{50}\times 6.60=46.67[/tex]

(a)

Compute the probability that he will earn at least $600 as follows:

[tex]P(S\geq 600)=P(\frac{S-\mu_{S}}{\sigma_{S}}\geq \frac{600-530}{46.67})\\\\=P(Z>1.50)\\\\=1-P(Z<1.50)\\\\=1-0.93319\\\\=0.06681\\\\\approx 0.0668[/tex]

Thus, the probability that he will earn at least $600 is 0.0668.

(b)

Let x represents his earnings on the best 1%  of such weekends.

That is, P (X < x) = 0.99.

⇒ P (Z < z) = 0.99

The corresponding z-score is, 2.33.

Compute the value of x as follows:

[tex]z=\frac{S-\mu_{S}}{\sigma_{S}}\\\\2.33=\frac{x-530}{46.67}\\\\x=530+(2.33\times 46.67)\\\\x=638.7411\\\\x\approx 638.74[/tex]

Thus, on the best 1%  of such weekends the waiter earned $638.74.


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Step-by-step explanation:

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In point 3:

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Answers

Answer:

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Step-by-step explanation:

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OR

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Answers

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Step-by-step explanation:

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b. It was expected thatâ 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is notâ 25%, do the results contradictâ expectations?

Answers

Answer:

a

The 95% confidence interval is [tex] 0.2474 < p < 0.3188 [/tex]

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  The result obtained does not contradict expectation

Step-by-step explanation:

From the question we are told that

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   The number of yellow peas is u =  173

Generally the sample size is mathematically represented as

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      [tex]\^ p = \frac{u}{n}[/tex]

=>   [tex]\^ p = \frac{173}{611 }[/tex]

=>   [tex]\^ p = 0.2831[/tex]

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=>   [tex]\alpha = 0.05[/tex]

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=>  [tex]E =  1.96 * \sqrt{\frac{ 0.2831 (1- 0.2831)}{611} } [/tex]

=>  [tex]E =  0.0357 [/tex]

Generally 95% confidence interval is mathematically represented as  

      [tex]\^ p -E <  p <  \^ p +E[/tex]

=>     [tex] 0.2831  -0.0357 <p< 0.2831  + 0.0357[/tex]

=>     [tex] 0.2474 < p < 0.3188 [/tex]

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Answers

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Step-by-step explanation:

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Answers

Answer:

The answer is C because you have to multiply them both:)

Answer:

The answer is C

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5 : 1 * 5

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Answers

Answer:

27

Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

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