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3 years ago

1. Determine the area under the standard normal curve to the left of z=2.10.

1 answer:

Nadusha1986 [10]3 years ago

3 0

The answer is 0.982

In your calculator you need to go to 2nd distr and then normalcdf. Then enter the lower bound as -99, the upper bound as 2.1, mean as 0 and standard deviation as 1. Then press enter and you should get 0.982 as your answer :)

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A researcher wants to estimate the percentage of all adults that have used the Internet to seek pre-purchase information in the

Lubov Fominskaja [6]

Answer:

The required sample size for the new study is 801.

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:

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

25% of all adults had used the Internet for such a purpose

This means that $\pi = 0.25$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

What is the required sample size for the new study?

This is n for which M = 0.03. So

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

$0.03 = 1.96\sqrt{\frac{0.25*0.75)}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.25*0.75}$

$\sqrt{n} = \frac{1.96\sqrt{0.25*0.75}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.25*0.75}}{0.03})^2$

$n = 800.3$

Rounding up:

The required sample size for the new study is 801.

4 0

3 years ago

The perimeter of the new m-phone is 16 inches. The phones length is 5.25 inches. What is the width of the phone

True [87]

Perimeter is all the sides added up together.
Divide 16 by 2, so that you can get just a measurement of the length and the width.

16/2 = 8

Since the length is 5.25, subtract that by 8.

8-5.25 = 2.75

The width of the phone is 2.75 inches.

5 0

3 years ago

Can someone help pls

lora16 [44]

I got 7.1 x 10^(-11) but it's been a few years so I could've done it wrong