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Semenov [28]
3 years ago
12

A national college researcher reported that 67%of students who graduated from high school in 2012 enrolled in college. Twenty-ni

ne high school graduates are sampled What is the mean number who enroll in college in a sample of 29 high school graduates? Round the answer to two decimal places. a) What is the standard deviation of the number who enroll in college in a sample of 29 high school graduates? Round the answer to four decimal places.
Mathematics
1 answer:
mrs_skeptik [129]3 years ago
7 0

Answer:

Mean is 19.43

Standard deviation is 2.5322

Step-by-step explanation:

For each student who graduated from high school in 2012, there are only two possible outcomes. Either they have enrolled in college, or they have not. The probability of a student having enrolled in college is independent of other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

67% of students who graduated from high school in 2012 enrolled in college.

This means that p = 0.67

Sample of 29 high school graduates

This means that n = 29

Mean

E(X) = np = 29*0.67 = 19.43

Standard deviation

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{29*0.67*0.33} = 2.5322

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Pilar is playing with a motorized toy boat. She puts the boat in a lake and it travels 400m at a constant speed. On the way back
ankoles [38]

Answer:

The trip normally takes 8 minutes

Step-by-step explanation:

The given information states that the away distance the boat traveled = 400 m

The time traveled at the same initial  speed , v₁, by the boat on the way back = 2 minutes

The increase in speed of the boat by Pilar =  10 m/min

The new speed, v₂ = v₁ + 10

The time for the return trip, t₂ = 60 seconds (1 minute) faster than time for the trip, t₁

t₂ = t₁ - 1

Therefore we have;

v₁ × t₁ = v₁×2 + v₂×(t₂-2) = 400

v₁×2 + (v₁ + 10)×(t₂-2) = 400

(v₁ + 10)×t₂ - 20 = 400

But v₁ = 400/t₁ = 400/(t₂ + 1)

Which gives;

(400/(t₂ + 1) + 10)×t₂ - 20 = 400

10×(t₂²+ 36·t₂-2)/(t₂+1) = 400

10·t₂²+ 10·t₂-420 = 0

t₂²+ t₂-42 = 0

(t₂ - 7)(t₂ + 6) = 0

t₂ = 7 minutes or -6 minutes

Given that t₂ is a natural number, we have, t₂ = 7 minutes

Whereby, t₂ = t₁ - 1, we have;

7 =  t₁ - 1

t₁ = 1 + 7 = 8 Minutes

The trip normally takes 8 minutes

3 0
3 years ago
What is the value of "?" if the Perimeter = 50 km?
777dan777 [17]

Answer:

14km

Explanation

11km + 11km = 22km is the sum of both the width.

If you want to find the length than minus the perimeter with 22.

50km - 22km = 28km is the sum of both the length.

To know only one side of the length, divide it by 2.

28km/2 = 14km is the length

6 0
2 years ago
you run a business making birdhouses. you spend $600 to start your business, and it costs you $5.00 to make each birdhouse.
Alex787 [66]

Answer:

???

Step-by-step explanation:

What is the question?

4 0
3 years ago
Graph the line whose x-intercept is 8 and whose y-intercept is -9.
eduard

Answer: attached image

7 0
1 year ago
A random sample of n 1 = 249 people who live in a city were selected and 87 identified as a democrat. a random sample of n 2 = 1
kvasek [131]

Answer:

CI=\{-0.2941,-0.0337\}

Step-by-step explanation:

Assuming conditions are met, the formula for a confidence interval (CI) for the difference between two population proportions is \displaystyle CI=(\hat{p}_1-\hat{p}_2)\pm z^*\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2(1-\hat{p}_2)}{n_2} where \hat{p}_1 and n_1 are the sample proportion and sample size of the first sample, and \hat{p}_2 and n_2 are the sample proportion and sample size of the second sample.

We see that \hat{p}_1=\frac{87}{249}\approx0.3494 and \hat{p}_2=\frac{58}{113}\approx0.5133. We also know that a 98% confidence level corresponds to a critical value of z^*=2.33, so we can plug these values into the formula to get our desired confidence interval:

\displaystyle CI=(\hat{p}_1-\hat{p}_2)\pm z^*\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\\\\CI=\biggr(\frac{87}{249}-\frac{58}{113}\biggr)\pm 2.33\sqrt{\frac{\frac{87}{249}(1-\frac{87}{249})}{249}+\frac{\frac{58}{113}(1-\frac{58}{113})}{113}}\\\\CI=\{-0.2941,-0.0337\}

Hence, we are 98% confident that the true difference in the proportion of people that live in a city who identify as a democrat and the proportion of people that live in a rural area who identify as a democrat is contained within the interval {-0.2941,-0.0337}

The 98% confidence interval also suggests that it may be more likely that identified democrats in a rural area have a greater proportion than identified democrats in a city since the differences in the interval are less than 0.

5 0
2 years ago
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