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lyudmila [28]
3 years ago
12

The Census Bureau reports that 82% of Americans over the age of 25 are high school graduates. A survey of randomly selected resi

dents of certain county included 1110 who were over the age of 25, and 871 of them were high school graduates.Find the mean and standard deviation for the number of high school graduates in groups of 1110 Americans over the age of 25.
Mathematics
1 answer:
mihalych1998 [28]3 years ago
6 0

Answer:

Mean = 910.2

Standard deviation = 12.8

Step-by-step explanation:

82% over the age of 25 are high school graduates.

A survey randomly selected students of certain country included 1110 who were over the age of 25 and 871 of them were high school.

n = 1110

p = 82% = 0.82

q = 1 - 0. 82

q = 0.18

Mean = np

= 1110*0.82

= 910.2

Standard deviation = √npq

= √ 1110*0.82*0.18

=√163.836

= 12.8

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Answer:

Step-by-step explanation:

Gabrielhs 59 flowers because 43 + 16= 59.

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A driving exam consists of 30 ​multiple-choice questions. Each of the 30 answers is either right or wrong. Suppose the probabili
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Answer:

a. P(x>20)=0.19

b. P(x≥6)=0.72

c. P(x≤20)=0.81

d. A and C

Step-by-step explanation:

We know that:

1) the probability that a student makes fewer than 6 mistakes is 0.28

P(x

2) The probaiblity that a student makes between 6 to 20 mistakes is 0.53.

P(6\leq x\leq20)=0.53

We will express the proabilibities in function of the information we have.

a. Probability that a student makes more than 20 mistakes.

P(x>20)=1-P(x\leq20)=1-(P(x20)=1-(0.28+0.53)=1-0.81=0.19

b. Probability that the student make 6 or more mistakes

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c. Probability that a student makes 20 mistakes at most

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d. A and C, because A takes a event of more than 20 mistakes and C takes the event of 20 or less mistakes. Both events cover a probability of 1.

7 0
3 years ago
4{3sqaured+4(116-8(2×8))} I need to know ASAP it 9:30 where I live pls hwlp
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3 years ago
A random sample of 864 births in a state included 426 boys. Construct a 95% confidence interval estimate of the proportion of bo
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<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

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

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, we have a 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 the critical value is z = 1.96.

We have that a random sample of 864 births in a state included 426 boys, hence the parameters are given by:

n = 864, \pi = \frac{426}{864} = 0.493

Then, the bounds of the interval are given by:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.493 - 1.96\sqrt{\frac{0.493(0.507)}{864}} = 0.46

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.493 + 1.96\sqrt{\frac{0.493(0.507)}{864}} = 0.526

The 95% confidence interval estimate of the proportion of boys in all births is (0.46, 0.526). Since the interval contains 0.506, it does not provide strong evidence against that belief.

More can be learned about the z-distribution at brainly.com/question/25890103

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2 years ago
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Step-by-step explanation:

the solution is (0.25,-0.5) because when you graph them on a graph, they both intercept at that point.

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3 years ago
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