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Leto [7]
3 years ago
11

Tom surveyed a random sample of the junior of his school to determine whether the Fall Festival should be held in October or Nov

ember. Of the 80 students surveyed, 24.8% said they preferred November. Based on this information, about how many students in the entire 230-person class would be expected to prefer having the Fall Festival in November. SHOW YOUR WORK PLEASE!!!
a. 50
b. 60
c. 75
d. 80
Mathematics
1 answer:
san4es73 [151]3 years ago
3 0

9514 1404 393

Answer:

  b.  60

Step-by-step explanation:

We assume the percentage for the sample holds for the whole class, so the estimated number preferring November is ...

  0.248 × 230 = 57.04 ≈ 60

About 60 students prefer November.

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Its only divisible by 2.
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3 years ago
The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centime
Minchanka [31]

Answer:

Step-by-step explanation:

Hello!

For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.

In this example the variable is:

X: height of a college student. (cm)

There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.

The option you have is to apply the Central Limit Theorem.

The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:

X[bar]~~N(μ;σ2/n)

Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:

98% CI

1 - α: 0.98

⇒α: 0.02

α/2: 0.01

Z_{1-\alpha /2}= Z_{1-0.01}= Z_{0.99} =2.334

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

174.5 ± 2.334* \frac{6.9}{\sqrt{50} }

[172.22; 176.78]

With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].

I hope it helps!

4 0
3 years ago
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never [62]

Answer:

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domain is a set of all x values, so just write the x values based on the graph

range is a set of all y values, so just write y values based on the graph

not function because it doesnt satisfy with the vertical line test.

7 0
3 years ago
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Ilia_Sergeevich [38]

Answer:

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Step-by-step explanation:

we know that

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8 0
3 years ago
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