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Stels [109]
3 years ago
10

Let X1, X2, and X3 represent the times necessary to perform three successive repair tasks at a service facility. Suppose they ar

e normal random variables with means of 50 minutes, 60 minutes, and 40 minutes, respectively. The standard deviations are 15 minutes, 20 minutes, and 10 minutes, respectively. a Suppose X1, X2, and X3 are independent. All three repairs must be completed on a given object. What is the mean and variance of the total repair time for this object.
Mathematics
1 answer:
belka [17]3 years ago
8 0

Answer:

The mean of the total repair time is 150 minutes.

The variance of the total repair time is 725 minutes^2.

Step-by-step explanation:

To solve this problem, we have to use the properties of the mean and the variance. Our random variable is the sum of 3 normal variables.

In the case, for the mean, we have that the mean of the sum of 3 normal variables is equal to the sum of the mean of the 3 variables:

y=x_1+x_2+x_3 \\\\E(y)=E(x_1+x_2+x_3)=E(x_1)+E(x_2)+E(x_3)\\\\E(y)=50+60+40=150

For the variance, we apply the property for the sum of independent variables (the correlation between the variables is 0):

V(y)=V(x_1)+V(x_2)+V(x_3)\\\\V(y)=s_1^2+s_2^2+s_3^2\\\\V(y)=15^2+20^2+10^2\\\\V(y)=225+400+100\\\\V(y)=725

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(9x + 8y2) • (9x - 8y2)

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dawn and pablo are members of a normally distributed population that is being sampled. if the chance of dawn being included in t
pav-90 [236]
Pablo must have the same chance as Dawn of being included in the sample, in order for valid inferences to be made based on the sample.
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4 0
3 years ago
Read 2 more answers
The average cost of tuition plus room and board at small private liberal arts colleges is reported to be less than $18,500 per t
lys-0071 [83]

Answer:

The null and alternative hypothesis for this study are:

H_0: \mu=18500\\\\H_a:\mu< 18500

The null hypothesis is rejected (P-value=0.004).

There is enough evidence to support the claim that the average cost of tuition plus room and board at small private liberal arts colleges is  less than $18,500 per term.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the average cost of tuition plus room and board at small private liberal arts colleges is  less than $18,500 per term.

Then, the null and alternative hypothesis are:

H_0: \mu=18500\\\\H_a:\mu< 18500

The significance level is 0.05.

The sample has a size n=150.

The sample mean is M=18200.

The standard deviation of the population is known and has a value of σ=1400.

We can calculate the standard error as:

\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{1400}{\sqrt{150}}=114.31

Then, we can calculate the z-statistic as:

z=\dfrac{M-\mu}{\sigma_M}=\dfrac{18200-18500}{114.31}=\dfrac{-300}{114.31}=-2.624

This test is a left-tailed test, so the P-value for this test is calculated as:

P-value=P(z

As the P-value (0.004) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the average cost of tuition plus room and board at small private liberal arts colleges is  less than $18,500 per term.

3 0
4 years ago
Lucy made $132 for 11 hours of work.
OlgaM077 [116]

Answer:

132/11 = 12

204/12 = 17

17 hours

4 0
3 years ago
2 4/5 x 1/3 = <br> 1 1/4 x 1/5 =
g100num [7]

Answer:

\frac{14}{15}
\frac{1}{4}

Step-by-step explanation:

Given the following equations:

2\frac{4}{5} \times\frac{1}{3}
1\frac{1}{4} \times\frac{1}{5}

In order to solve the following equations, we convert the mixed numbers into improper fractions and then solve by multiplying the numerators and the denominators by each other.

2\frac{4}{5} \times\frac{1}{3}
2\frac{4}{5} =5\times2=10+4=\frac{14}{5}
\frac{14}{5} \times\frac{1}{3}
14\times1=14
5\times3=15
=\frac{14}{15}

1\frac{1}{4} \times\frac{1}{5}
1\frac{1}{4} =1\times4=4+1=\frac{5}{4}
\frac{5}{4} \times\frac{1}{5}
5\times1=5
4\times5=20
\frac{5}{20} \div5=\frac{1}{4}
=\frac{1}{4}

Hope this helps.

6 0
2 years ago
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