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gayaneshka [121]
3 years ago
15

The mean cost of a five pound bag of shrimp is 46 dollars with a variance of 64. If a sample of 53 bags of shrimp is randomly se

lected, what is the probability that the sample mean would differ from the true mean by greater than 2.1 dollars? Round your answer to four decimal places.
Mathematics
1 answer:
sertanlavr [38]3 years ago
6 0

Answer:

The probability that sample mean differ the true greater than 2.1 will be 2.8070 %

Step-by-step explanation:

Given:

Sample mean =46 dollars

standard deviation=8

n=53

To Find :

Probability that sample mean would  differ from true mean by greater than 2.1

Solution;

<em>This sample distribution mean problem,</em>

so for that

calculate Z- value

Z=(sample mean - true mean)/(standard deviation/Sqrt(n))

Z=-2.1/(8/Sqrt(53))

Z=-2.1*Sqrt(53)/8

Z=-1.91102

Now for P(X≥2.1)=P(Z≥-1.91102)

Using Z-table,

For Z=-1.91

P(X>2.1)=0.02807

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For this exercise it is important to remember the following:

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1. Replace \sqrt{-1} with i and simplify:

(-2i)(i)\sqrt{12}=-2i^2\sqrt{12}=-2(-1)\sqrt{12}=2\sqrt{12}

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One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst
grandymaker [24]

Answer:

(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(c) P (X = 3) = 0.09.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let <em>X</em> = number of telephone calls.

The average number of calls per minute is, <em>λ</em> = 3.0.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 3.0.

The probability mass function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...

(a)

Compute the probability of <em>X</em> = 0 as follows:

P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498

Thus, the  probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of <em>X</em> > 5  as follows:

P (X > 5) = 1 - P (X ≤ 5)

              =1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of <em>X</em> = 3 as follows:

<em> </em>P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09<em />

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of <em>X</em> ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

             =\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

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