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Kipish [7]
3 years ago
13

A research study shows that the yearly cost of dental claims for virginia state employees with dental care as a part of their he

alth insurance has a distribution with a mean of $1200 and a standard deviation of $200. if a random sample of 25 virginia state employees with dental insurance is selected, what is the probability that the mean of the yearly dental claims for this sample is between $1175 and $1225?
Mathematics
1 answer:
Musya8 [376]3 years ago
3 0
The right answer is 50% .
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Akron Cinema sells an average of 500 tickets on Mondays, with a standard deviation of 50 tickets. If a simple random sample is t
Paha777 [63]

Answer:

Step-by-step explanation:

Assuming the number of tickets sales from Mondays is normally distributed. the formula for normal distribution would be applied. It is expressed as

z = (x - u)/s

Where

x = ticket sales from monday

u = mean amount of ticket

s = standard deviation

From the information given,

u = 500 tickets

s = 50 tickets

We want to find the probability that the mean will be greater than 510. It is expressed as

P(x greater than 510) = 1 - P(x lesser than or equal to 510)

For x = 510

z = (510 - 500)/50 = 0.2

Looking at the normal distribution table, the probability corresponding to the z score is 0.9773

P(x greater than 510) = 1 - 0.9773 = 0.0227

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2 years ago
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1 1/5 as decimal because im doing surface area but its hard​
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Answer:

1.2

Step-by-step explanation:

1 \frac{1}{5\\}=1.2

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3 years ago
Write the place value of each decimals 1.0 59​
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Answer:

1.059×1000 is the value of the following

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A human resources representative claims that the proportion of employees earning more than $50,000 is less than 40%. To test thi
bearhunter [10]

Answer:

The statistic for this case would be:

z=\frac{\hat p -p_o}{\sqrt{\frac{\hat p(1-\hat p)}{n}}}

And replacing we got:

z= \frac{0.436-0.4}{\sqrt{\frac{0.436*(1-0.436)}{700}}}= 1.92

Step-by-step explanation:

For this case we have the following info:

n =700 represent the sample size

X= 305 represent the number of employees that earn more than 50000

\hat p=\frac{305}{700}= 0.436

We want to test the following hypothesis:

Nul hyp. p \leq 0.4

Alternative hyp : p>0.4

The statistic for this case would be:

z=\frac{\hat p -p_o}{\sqrt{\frac{\hat p(1-\hat p)}{n}}}

And replacing we got:

z= \frac{0.436-0.4}{\sqrt{\frac{0.436*(1-0.436)}{700}}}= 1.92

And the p value would be given by:

p_v = P(z>1.922)= 0.0274

4 0
3 years ago
in 1981, about 1.1 million was lost due to fires. In 1988, The loss was about 9.6 million. Find each percentage of change
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