Catherine went bowling. She rented

A college-entrance exam is designed so that scores are normally distributed with a mean of 500 and a standard deviation of 100.

yawa3891 [41]

Answer:

The probability is 0.003

Step-by-step explanation:

We know that the average $\mu$ is:

$\mu=500$

The standard deviation $\sigma$ is:

$\sigma=100$

The Z-score is:

$Z=\frac{x-\mu}{\sigma}$

We seek to find

$P(x800)$

For P(x>800) The Z-score is:

$Z=\frac{x-\mu}{\sigma}$

$Z=\frac{800-500}{100}$

$Z=3$

The score of Z = 3 means that 800 is 3 standard deviations from the mean. Then by the rule of the 8 parts of the normal curve, the area that satisfies the conficion of 3 deviations from the mean has percentage of 0.15%

So

$P(x>800)=0.15\%$

For P(x<200) The Z-score is:

$Z=\frac{x-\mu}{\sigma}$

$Z=\frac{200-500}{100}$

$Z=-3$

The score of Z = -3 means that 200 is 3 standard deviations from the mean. Then by the rule of the 8 parts of the normal curve, the area that satisfies the conficion of 3 deviations from the mean has percentage of 0.15%

So

$P(x$

Therefore

$P(x800)=P(x800)$

$P(x800)=0.0015 + 0.0015$

$P(x800)=0.003$

6 0

3 years ago

Can y’all help me???

Andreyy89

You will need 3 cups :)

6 0

3 years ago

Which figure shows that

FromTheMoon [43]

It’s either Option 1 or 4

7 0

3 years ago

Read 2 more answers

2x - 1/2 Płēâšė hęł hęłp mè ï ńèèd thîš før hōmèwörk

Lynna [10]

Answer for x

2x-1/2=
2x=1/2 multiply both sides with 2
4x=1
x=1/4

7 0

3 years ago

A fitness center is interested in finding a 90% confidence interval for the mean number of days per week that Americans who are

Blababa [14]

Answer:

A. Normal

B. Between 2.078 and 2.722 visits.

C. About 90 percent of these confidence intervals will contain the true population mean number of visits per week and about 10 percent will not contain the true population mean number of visits per week.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

x% confidence interval:

A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.

Question A:

By the Central Limit Theorem, normal.

Question B:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.05 = 0.95$ , so Z = 1.645.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.645\frac{2.9}{\sqrt{219}} = 0.322$

The lower end of the interval is the sample mean subtracted by M. So it is 2.4 - 0.322 = 2.078 visits

The upper end of the interval is the sample mean added to M. So it is 2.4 + 0.322 = 2.722 visits.

Between 2.078 and 2.722 visits.

Question c:

90% confidence level, so 90% will contain the true population mean, 100 - 90 = 10% wont.

About 90 percent of these confidence intervals will contain the true population mean number of visits per week and about 10 percent will not contain the true population mean number of visits per week.

5 0

3 years ago