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PolarNik [594]
2 years ago
12

When a production process is operating correctly, thenumber of units produced per hour has a normal distri- bution with a mean o

f 92.0 and a standard deviation of3.6. A random sample of 4 different hours was taken.a. Find the mean of the sampling distribution of thesample means. b. Find the variance of the sampling distribution ofthe sample mean.c. Find the standard error of the sampling distribu-tion of the sample mean.d. What is the probability that the sample mean ex-ceeds 93.0 units?
Mathematics
1 answer:
KengaRu [80]2 years ago
3 0

Answer:

a) The mean of the sampling distribution of the sample means is 92.

b) The variance of the sampling distribution of the sample mean is 3.24.

c) The standard error of the sampling distribution of the sample mean is 1.8.

d) 28.77% probability that the sample mean ex-ceeds 93.0 units.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 92, \sigma = 3.6.

a. Find the mean of the sampling distribution of thesample means.

The mean is the same as the mean of the population. So the mean of the sampling distribution of the sample means is 92.

b. Find the variance of the sampling distribution ofthe sample mean.

The standard deviation is s = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{\sqrt{4}} = 1.8

The variance is s^{2} = 1.8^{2} = 3.24

c. Find the standard error of the sampling distribution of the sample mean.

This is the same as the standard deviation of the sample. So the standard error of the sampling distribution of the sample mean is 1.8.

d. What is the probability that the sample mean ex-ceeds 93.0 units?

This is 1 subtracted by the pvalue of Z when X = 93. So

Z = \frac{X - \mu}{s}

Z = \frac{93 - 92}{1.8}

Z = 0.56

Z = 0.56 has a pvalue of 0.7123.

So there is a 1-0.7123 = 0.2877 = 28.77% probability that the sample mean ex-ceeds 93.0 units.

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Answer:

b. 0.864

Step-by-step explanation:

Let's start defining the random variables.

Y1 : ''Number of customers who spend more than $50 on groceries at counter 1''

Y2 : ''Number of customers who spend more than $50 on groceries at counter 2''

If X is a binomial random variable, the probability function for X is :

P(X=x)=(nCx)p^{x}(1-p)^{n-x}

Where P(X=x) is the probability of the random variable X to assume the value x

nCx is the combinatorial number define as :

nCx=\frac{n!}{x!(n-x)!}

n is the number of independent Bernoulli experiments taking place

And p is the success probability.

In counter I :

Y1 ~ Bi (n,p)

Y1 ~ Bi(2,0.2)

P(Y1=y1)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}

With y1 ∈ {0,1,2}

And P( Y1 = y1 ) = 0 with y1 ∉ {0,1,2}

In counter II :

Y2 ~ Bi (n,p)

Y2 ~ Bi (1,0.3)

P(Y2=y2)=(1Cy2)(0.3)^{y2}(0.7)^{1-y2}

With y2 ∈ {0,1}

And P( Y2 = y2 ) = 0 with y2 ∉ {0,1}

(1Cy2) with y2 = 0 and y2 = 1 is equal to 1 so the probability function for Y2 is :

P(Y2=y2)=(0.3)^{y2}(0.7)^{1-y2}

Y1 and Y2 are independent so the joint probability distribution is the product of the Y1 probability function and the Y2 probability function.

P(Y1=y1,Y2=y2)=P(Y1=y1).P(Y2=y2)

P(Y1=y1,Y2=y2)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}(0.3)^{y2}(0.7)^{1-y2}

With y1 ∈ {0,1,2} and y2 ∈ {0,1}

P( Y1 = y1 , Y2 = y2) = 0 when y1 ∉ {0,1,2} or y2 ∉ {0,1}

b. Not more than one of three customers will spend more than $50 can mathematically be expressed as :

Y1 + Y2 \leq 1

Y1 + Y2\leq 1 when Y1 = 0 and Y2 = 0 , when Y1 = 1 and Y2 = 0 and finally when Y1 = 0 and Y2 = 1

To calculate P(Y1+Y2\leq 1) we must sume all the probabilities that satisfy the equation :

P(Y1+Y2\leq 1)=P(Y1=0,Y2=0)+P(Y1=1,Y2=0)+P(Y1=0,Y2=1)

P(Y1=0,Y2=0)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{0}(0.7)^{1-0}=(0.8)^{2}(0.7)=0.448

P(Y1=1,Y2=0)=(2C1)(0.2)^{1}(0.8)^{2-1}(0.3)^{0}(0.7)^{1-0}=2(0.2)(0.8)(0.7)=0.224

P(Y1=0,Y2=1)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{1}(0.7)^{1-1}=(0.8)^{2}(0.3)=0.192

P(Y1+Y2\leq 1)=0.448+0.224+0.192=0.864\\P(Y1+Y2\leq 1)=0.864

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Answer:

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b) The expected number of days is 10 days.

Step-by-step explanation:

The most appropiate distribution to use in this case is the geometric distribution, in order to calculate the probability of a success after k failure trials.

The probability of success, as each of the 10 products are assumed to have fair probabilities, is:

p=1/10=0.1

Then, the probability that our product is not selected any given day is:

q=1-p=1-0.1=0.9

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P=q^9p^1=0.9^9\cdot0.1=0.3874\cdot0.1=0.039

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General Idea:

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In the same way if P(x, y) moves right or left ' h ' units, then we can represent that transformation by adding or subtracting respectively 'h' units to the x-coordinate.

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