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wariber [46]
3 years ago
10

Someone vibe with me<3 (look at comments)

Mathematics
1 answer:
Akimi4 [234]3 years ago
4 0

<em>Answer:</em>

<em>I will vibe with you my friend :3</em>

<em />

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Answer:

Case n =5

z=\frac{4.8-5}{\frac{0.5}{\sqrt{5}}}=-0.894  

Case n =15

z=\frac{4.8-5}{\frac{0.5}{\sqrt{15}}}=-1.549  

Case n = 40

z=\frac{4.8-5}{\frac{0.5}{\sqrt{40}}}=-2.530  

P value

Case n =5

p_v =P(z  

Case n =15

p_v =P(z  

Case n =40

p_v =P(z  

Step-by-step explanation:

Data given and notation  

\bar X=4.8 represent the sample mean

\sigma=0.5 represent the population standard deviation

n sample size  

\mu_o =5 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is lower than 5, the system of hypothesis would be:  

Null hypothesis:\mu \geq 5  

Alternative hypothesis:\mu < 5  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

Case n =5

z=\frac{4.8-5}{\frac{0.5}{\sqrt{5}}}=-0.894  

Case n =15

z=\frac{4.8-5}{\frac{0.5}{\sqrt{15}}}=-1.549  

Case n = 40

z=\frac{4.8-5}{\frac{0.5}{\sqrt{40}}}=-2.530  

P-value  

Since is a left tailed test the p value would be:  

Case n =5

p_v =P(z  

Case n =15

p_v =P(z  

Case n =40

p_v =P(z  

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Step-by-step explanation:

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Do not round intermediate calculations. round your answers to two decimal places. a. define a random variable number of times th
navik [9.2K]

Observations can you make from a comparison of the number of water supply stoppages reported by owner-occupied units versus renter-occupied units 1.

Total no. of times owner-occupied units had a water supply stoppage lasting 6 or more hours

547000 + 5012000 + 6110000 + 2544000 + 557000 = 14770000

x = no. of times owner-occupied units had a water supply stoppage.

<h3>What is the probability at x =0?</h3>

P(x=0) = 547000/14770000

P(x=0) = 0.0370

Similarly, we have at x=1

P(x=1) = 5012000/14770000

P(x=1) = 0.3393

P(x=2) = 6110000/14770000

P(x=2) = 0.4137

P(x=3) = 2544000/14770000

P(x=3) = 0.1722

P(x=4) = 557000/14770000

P(x=4) = 0.0378

x           f(x)

0             0.0370

1              0.3393

2         0.4137

3              0.1722

4         0.0378

Total     1

Observations can you make from a comparison of the number of water supply stoppages reported by owner-occupied units versus renter-occupied units 1.

To learn more about the probability distribution visit:

brainly.com/question/24756209

#SPJ1

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