How many times can 96 go in 1000?

A programmer plans to develop a new software system. In planning for the operating system that he will use, he needs to estimat

Vedmedyk [2.9K]

Using the z-distribution, we have that:

a) A sample of 601 is needed.

b) A sample of 93 is needed.

c) A. Yes, using the additional survey information from part (b) dramatically reduces the sample size.

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

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

In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

The margin of error is:

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

95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

For this problem, we consider that we want it to be within 4%.

Item a:

The sample size is <u>n for which M = 0.04.</u>

There is no estimate, hence $\pi = 0.5$

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

$0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}$

$\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2$

$n = 600.25$

Rounding up:

A sample of 601 is needed.

Item b:

The estimate is $\pi = 0.96$ , hence:

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

$0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}$

$\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2$

$n = 92.2$

Rounding up:

A sample of 93 is needed.

Item c:

The closer the estimate is to $\pi = 0.5$ , the larger the sample size needed, hence, the correct option is A.

For more on the z-distribution, you can check brainly.com/question/25404151

8 0

2 years ago

how would you change the sentence into a true statement? some irrational numbers are also rational numbers..... A. All irrationa

Fynjy0 [20]

Irrational numbers cannot be classified as rational numbers. - EXPLANATION - both are different completely, according to as I think, and thus they cannot be inter-related. Irrational like - root2, root3, and all those that cannot be further solved completely. Their decimal representation is neither recurring nor ending e.g. 1.101001000100001000001.... and so on while rational number are consisting of whole numbers, integers, natural numbers, and all the constants and their decimal representation can be non ending but recurring.

5. d. Irrational numbers cannot be classified as rational numbers. - EXPLANATION - out of the options this one's correct. As I above said, all numbers as far as I know can be categorized into unreal and real numbers out of which real numbers consist of irrational and rational numbers while unreal numbers, I don't know as I have not been taught about them yet and neither do you have been, as I think so. All I know is that, as far as I have been taught, unreal number are not connected to real numbers.

6. Every real number is a rational number. - EXPLANATION - As I told you above, real numbers can also be irrational so this is false.

Hope this helps!

8 0

3 years ago

Read 2 more answers

According to a study done by a university student, the probability a randomly selected individual will not cover his or her mou

Sergio [31]

Using the binomial distribution, it is found that:

a) There is a 0.1618 = 16.18% probability that among 18 randomly observed individuals exactly 6 do not cover their mouth when sneezing.

b) There is a 0.104 = 10.4% probability that among 18 randomly observed individuals fewer than 3 do not cover their mouth when sneezing.

c) 9 is more than 2.5 standard deviations below the mean, hence it would not be surprising if fewer than half covered their mouth when sneezing.

<h3>What is the binomial distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

The values of the parameters are given as follows:

n = 18, p = 0.267.

Item a:

The probability is P(X = 6), hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{18,6}.(0.267)^{6}.(0.733)^{12} = 0.1618$

There is a 0.1618 = 16.18% probability that among 18 randomly observed individuals exactly 6 do not cover their mouth when sneezing.

Item b:

The probability is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2).

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{18,0}.(0.267)^{0}.(0.733)^{18} = 0.0037$

$P(X = 1) = C_{18,1}.(0.267)^{1}.(0.733)^{17} = 0.0245$

$P(X = 2) = C_{18,2}.(0.267)^{2}.(0.733)^{16} = 0.0758$

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0037 + 0.0245 + 0.0758 = 0.104.

There is a 0.104 = 10.4% probability that among 18 randomly observed individuals fewer than 3 do not cover their mouth when sneezing.

item c:

We have to look at the mean and the standard deviation, given, respectively, by:

E(X) = np = 18 x 0.267 = 4.81.

$\sqrt{V(X)} = \sqrt{18(0.267)(0.733)} = 1.88$

9 is more than 2.5 standard deviations below the mean, hence it would not be surprising if fewer than half covered their mouth when sneezing.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

6 0

2 years ago

Angel had $ 123.47 to begin her trip she went a little wild d spent $8756 during the first three days f the trip. She still has

Anna [14]

87.56????????
Just subtract 87.56 from 123.47 then divide by 2 .
= 17.95

3 0

3 years ago

Read 2 more answers

(please explain each step) <br><br>Find the m<QRP if m<PRS=35° and m<QRS=85°

iragen [17]

You have to find x (QRP), so just subtract the entire angle(QRS) with the given angle (PRS) to get your missing angle(QRP).

Hope this helps.

3 0

4 years ago