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grandymaker [24]
3 years ago
14

Find the expected value of the following probability distribution. X 1 2 3 4 5 P(X) 0.05 0.20 0.35 0.25 0.15 Round your answer t

o two decimal place as necessary. For example, 4.56 would be a legitimate entry.Expected value = ___.
Mathematics
1 answer:
Phoenix [80]3 years ago
6 0

Answer:

5.63

Step-by-step explanation:

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) The National Highway Traffic Safety Administration collects data on seat-belt use and publishes results in the document Occupa
asambeis [7]

Answer:

We conclude that there is a difference in seat belt use.

Step-by-step explanation:

We are given that of 1,000 drivers 16-24 years old, 79% said they buckle up, whereas 924 of 1,100 drivers 25-69 years old said they did.

<u><em>Let </em></u>p_1<u><em> = population proportion of drivers 16-24 years old who buckle up .</em></u>

<u><em /></u>p_2<u><em> = population proportion of drivers 25-69 years old who buckle up .</em></u>

So, Null Hypothesis, H_0 : p_1-p_2 = 0      {means that there is no significant difference in seat belt use}

Alternate Hypothesis, H_A : p_1-p_2\neq 0      {means that there is a difference in seat belt use}

The test statistics that would be used here <u>Two-sample z proportion statistics;</u>

                     T.S. =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }  ~ N(0,1)

where, \hat p_1 = sample proportion of drivers 16-24 years old who buckle up = 79%

\hat p_2 = sample proportion of drivers 25-69 years old who buckle up = \frac{924}{1100} = 84%

n_1 = sample of 16-24 years old drivers = 1000

n_2 = sample of 25-69 years old drivers = 1100

So, <u><em>test statistics</em></u>  =  \frac{(0.79-0.84)-(0)}{\sqrt{\frac{0.79(1-0.79)}{1000}+\frac{0.84(1-0.84)}{1100} } }  

                              =  -2.946

The value of z test statistics is -2.946.

Since, in the question we are not given with the level of significance so we assume it to be 5%. <u><em>Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.</em></u><em> </em>

<em>Since our test statistics doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that there is a difference in seat belt use.

4 0
3 years ago
Lucas invests $12,500 in an account that earns 6.2% compound interest if Lucas does not make any additional deposits or withdraw
kow [346]
Assuming it is compounded monthly, Lucas will have $28,128,737.86 in his account after 16 months. 
3 0
3 years ago
Did I do this correctly and if any errors just tell me in the comments
yulyashka [42]
Looks good to me! ;)
5 0
3 years ago
Read 2 more answers
A sample of students is taken from the school's A honor roll. The school estimates that there are actually 360 students on the A
aliya0001 [1]

Answer:

360 - Y

Step-by-step explanation:

From the information provided in the question, we will understand that the question lacks detailed information.

Given that:

The total number of actual students on the A honor roll = 360

To find the number of students on the A honor roll of 8th graders;

Let denote Y to represent the total number of students who exist on the A honor roll except for the 8th graders.

Then:

The number of 8th graders = all number of students on are on the A honor roll - Y

The number of 8th graders = 360 - Y

If Y is known, then the number of the 8th graders can be fully determined.

5 0
3 years ago
3.<br> 3 cm<br> 11 am<br> 6 cm<br> 4.3 cm<br> 8 cm
Archy [21]
32cm yay I got it right
8 0
3 years ago
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