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malfutka [58]
3 years ago
5

expectation of 24.0 mm and a standard deviation of 0.3 mm. What are the expectation and variance of the length of the machine pa

rt
Mathematics
1 answer:
Elena L [17]3 years ago
4 0
Are you sure the question is written correctly? Sounds like the expectations would just be the same (24.0) and the variance is the square of the standard deviation (so it would be 0.09).
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we believe that 42% of freshmen do not visit their counselors regularly. For this year, you would like to obtain a new sample to
Maksim231197 [3]

Answer:

A sample of 1077 is required.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is of:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

42% of freshmen do not visit their counselors regularly.

This means that \pi = 0.42

98% confidence level

So \alpha = 0.02, z is the value of Z that has a pvalue of 1 - \frac{0.02}{2} = 0.99, so Z = 2.327.

How large of a sample size is required?

A sample size of n is required, and n is found when M = 0.035. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.035 = 2.327\sqrt{\frac{0.42*0.58}{n}}

0.035\sqrt{n} = 2.327\sqrt{0.42*0.58}

\sqrt{n} = \frac{2.327\sqrt{0.42*0.58}}{0.035}

(\sqrt{n})^2 = (\frac{2.327\sqrt{0.42*0.58}}{0.035})^2

n = 1076.8

Rounding up:

A sample of 1077 is required.

3 0
3 years ago
What are the mean, median, and mode for the following data set?
FrozenT [24]
Order the number:
15,34,71,75,75,78,78,78,82,90,90,94,100
Mode:78
Mean≈74.8
Median:78 is your final answer. Hope it help!
3 0
4 years ago
Read 2 more answers
A circle with radius 6 has a sector with a central angle of 48 what is the area of the sector
inn [45]

Answer:

≈ 15.08 units²

Step-by-step explanation:

The area (A) of the sector is calculated as

A = area of circle × fraction of circle

   = πr² × \frac{48}{360} ( r is the radius )

   = π × 6² × \frac{48}{360}

   = 36π ×\frac{48}{360} = π × \frac{48}{10} = \frac{48\pi }{10} ≈ 15.08

5 0
3 years ago
Read 2 more answers
Suppose in the population, the Anger-Out score for men is two points higher than it is for women. The population variances for m
tatuchka [14]

Answer:

a)= 2

b) 6.324

c) P= 0.1217

Step-by-step explanation:

a) The mean of the sampling distribution of X`1- X`2 denoted by ux`-x` = u1-u2 is equal to the difference between population means i.e = 2 ( given in the question)

b) The standard deviation of the sampling distribution of X`1- X`2 ( standard error of X`1- X`2) denoted by σ_X`1- X`2 is given by

σ_X`1- X`2 = √σ²/n1 +σ²/n2

Var ( X`1- X`2) = Var X`1 + Var X`2 =  σ²/n1 +σ²/n2

so

σ_X`1- X`2 =√20 +20 = 6.324

if the populations are normal the sampling distribution X`1- X`2 , regardless of sample sizes , will be normal with mean u1-u2 and variance σ²/n1 +σ²/n2.

Where as Z is normally distributed with mean zero and unit variance.

If we take X`1- X`2= 0 and u1-u2= 2  and standard deviation of the sampling distribution = 6.324 then

Z= 0-2/ 6.342= -0.31625

P(-0.31625<z<0)= 0.1217

The probability would be 0.1217

8 0
3 years ago
Answer all please for me !!!
Andreyy89

Answer:

Step-by-step explanation:

3 0
3 years ago
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