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SIZIF [17.4K]
2 years ago
3

Tutoring Services: The Community College Survey of Student Engagement reports that 46% of the students surveyed rarely or never

use peer or other tutoring resources. Suppose that in reality 40% of community college students never use tutoring services available at their college. In a simulation we select random samples from a population in which 40% do not use tutoring. For each sample we calculate the proportion who do not use tutoring. If we randomly sample 100 students from this population, the standard error is approximately 5%. Would it be unusual to see 46% who do not use tutoring in a random sample of 100 students?
Mathematics
1 answer:
miskamm [114]2 years ago
6 0

Answer:

No, this would not be unusual because 46% is only 1.2 standard errors from 40%

Step-by-step explanation:

Consider the provided information.

The standard error is approximately 5%

Standard error(SE)=0.05

As we know SE=\frac{\sigma}{\sqrt{n} }

Here x=46%=0.46 and μ=40%=0.40

Now use the formula: z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}

Substitute the respective values in the above formula.

z=\frac{0.46-0.4}{0.05}

z=1.2

Hence, this would not be unusual because 46% is only 1.2 standard errors from 40%

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agasfer [191]

Answer:

Center (4, -3)  r=3

Step-by-step explanation:

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6 0
2 years ago
Use the fact that the variance of a Poisson distribution is σ2=μ. The mean number of bankruptcies filed per hour by businesses i
lianna [129]

Answer:

Variance = 5

Standard deviation = 2.236

0.2650257

Step-by-step explanation:

For a Poisson distribution is σ² = μ.

Given that:

Mean , μ of bankruptcies files per hour = 5

μ = 5

For a Poisson distribution :

P(x = x) = (μ^x * e^-μ) / x!

The Variance and standard deviation :

Variance : σ² = μ = 5

Standard deviation = sqrt(variance) = sqrt(5) = 2.236

B.) Find the probability that at most three businesses will file bankruptcy in any given hour.

P( x ≤ 3) = P(0) + P(1) + P(2) + P(3)

P(x = 0) = (5^0 * e^-5) / 0! = 0.0067379

P(x = 1) = (5^1 * e^-5) / 1! = 0.0336897

P(x = 2) = (5^2 * e^-5) / 2! = 0.0842243

P(x = 3) = (5^3 * e^-5) / 3! = 0.1403738

0.0067379 + 0.0336897 + 0.0842243 + 0.1403738

= 0.2650257

6 0
2 years ago
Akeem and Jack are discussing their friend's new swimming pool. It is rectangular in shape. Akeem says he knows that it has a pe
schepotkina [342]

Answer:

The length of swimming pool is 10 yards.

Step-by-step explanation:

Given,

Perimeter= 90 feet

Width= 5 yards

As we have to find the length in yards, therefore, we will convert perimeter in yards also.

As we know,

1 Yard = 3 feet

Therefore,

Dividing the perimeter by 3 to convert it into yards.

90\ feet = \frac{90}{3} \ yards\\90\ feet=30\ yards

The formula of perimeter of rectangle is;

P= 2(l+w)

Putting values of perimeter and width;

30=2(l+5)\\30=2l+10\\2l=30-10\\2l=20\\l=\frac{20}{2} \\l=10

The length of swimming pool is 10 yards.

Keywords: Perimeter, linear equation

Learn more about linear equations at:

  • brainly.com/question/6500232
  • brainly.com/question/6554402

#LearnwithBrainly

4 0
2 years ago
60 randomly selected students were asked how many siblings were in their family. Let X = the number of pairs of siblings in the
Burka [1]

Answer:

1. mean=2.47

2. median=2

3. Sample Standard deviation=1.24

4. First quartile=2

5. Third quartile=3

Step-by-step explanation:

The necessary calculations for finding mean,median,standard deviation, 1st quartile and 3rd quartile are

sumf=13+22+15+6+3+0+1=60

sumfx=13*1+22*2+15*3+6*4+3*5+0*6+1*7=148

sumfx²=13*1²+22*2²+15*3²+6*4²+3*5²+0*6²+1*7²=456

x   cumulative frequency

1    13

1    23+22=35

3   35+15=50

4    50+6=56

5    56+3=59

6    59+0=59

7     59+1=60

1.

mean=sum(fx)/sumf

Here x is number of siblings and f is frequency

mean=148/60=2.4667

So, after rounding to two decimal places our mean is 2.47.

2.

Median=value of (n/2)th observation

Median=value of (60/2)th observation

Median=value of 30th observation

The cumulative frequency shows that the 30th observation corresponds to x=2 sibling. So, the median is 2.

3.

s=\sqrt{\frac{sumfx^{2} -\frac{(sumfx)^{2} }{n} }{n-1} }

s=\sqrt{\frac{456 -\frac{(148)^{2} }{60} }{59} }

s=\sqrt{\frac{456 -365.0667 }{59} }}

s=\sqrt{1.5412 }

s=1.2415=1.24

4.

1st quartile=Q1=((n+1)/4)th values of observation

Q1=((61)/4)th values of observation

Q1=(15.25)th values of observation

15.25th value also corresponds to x=2 siblings so,

Q1=2.

5.

3rd quartile=Q3=(3(n+1)/4)th values of observation

Q3=(3(61)/4)th values of observation

Q3=(45.75)th values of observation

45.75th value  corresponds to x=3 siblings so,

Q3=3

8 0
2 years ago
Winnie measured the length of her father's ranch four times and got four different distances. her measurements are shown in the
Maksim231197 [3]

(a) winnie first approximated each distance to the nearest hundredth.

\sqrt{60} =7.75

\frac{58}{8} = 7.25

7.3 = 7.33

7\frac{3}{5} = 7.60

Now we find the average of 4 numbers

\frac{7.75+7.25+7.33+7.60}{4}= \frac{29.93}{4} = 7.4825

Winnie's estimate = 7.4825

B) Winnie's father estimated the distance across his ranch to be the square root of 56 km

\sqrt{56} = 7.48331

Winne's father estimate - Winne's estimate = 7.48331 - 7.48225= 0.00106

Winne's father estimate is 0.00106 more than winne's estimate

5 0
2 years ago
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