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JulsSmile [24]
3 years ago
15

Consider a random variable X that is normally distributed. Complete parts​ (a) through​ (d) below. ​(This is a reading assessmen

t question. Be certain of your answer because you only get one attempt on this​ question.) ​
(a) If a random variable X is normally​ distributed, what will be the shape of the distribution of the sample​ mean?
a. Normal
b. Skewed right
c. Skewed left
d. Cannot be determined
​(b) If the mean of a random variable X is 40​, what will be the mean of the sampling distribution of the sample​ mean? mu Subscript x overbarequals 40
​(c) As the sample size n​ increases, what happens to the standard error of the​ mean?
A. The standard error of the mean increases
B. The standard error of the mean remains the same.
C. The standard error of the mean decreases.
(d) If the standard deviation of a random variable X is 10 and a random sample of size nequals15 is​ obtained, what is the standard deviation of the sampling distribution of the sample​ mean? sigma Subscript x overbarequals nothing ​(Type an exact​ answer, using radicals as​ needed.)
Mathematics
1 answer:
Mariulka [41]3 years ago
5 0

Answer:

a)  a. Normal

b) The mean of the sampling distribution of the sample mean is 40.

c) C. The standard error of the mean decreases.

d) The standard deviation of the sampling distribution of the sample​ mean is 2.58.

Step-by-step explanation:

(a) If a random variable X is normally​ distributed, what will be the shape of the distribution of the sample​ mean?

If the variable is normally distributed, then the sample mean is going to be normally distributed. So the correct answer is

a. Normal

​(b) If the mean of a random variable X is 40​, what will be the mean of the sampling distribution of the sample​ mean?

The mean of the sampling distribuion is the same, so it is 40.

​(c) As the sample size n​ increases, what happens to the standard error of the​ mean?

We have that:

S_{E} = \frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation. So, as the sample size n increases, the standard error of the mean decreases. The correct answer is

C. The standard error of the mean decreases.

(d) If the standard deviation of a random variable X is 10 and a random sample of size nequals15 is​ obtained, what is the standard deviation of the sampling distribution of the sample​ mean

We have that:

s = \frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{15}} = 2.58

The standard deviation of the sampling distribution of the sample​ mean is 2.58.

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Answer:

C) \frac{2z+15}{6x-12y}

E) \frac{7d+5}{15d^2+14d+3}

F) \frac{-7a-b}{6b-4a}

Step-by-step explanation:

C)

One is given the following equation

\frac{z+1}{x-2y}-\frac{2z-3}{2x-4y}+\frac{z}{3x-6y}

In order to simplify fractions, one must convert the fractions to a common denominator. The common denominator is the least common multiple between the given denominators. Please note that the denominator is the number under the fraction bar of a fraction. In this case, the least common multiple of the denominators is (6x-12y). Multiply the numerator and denominator of each fraction by the respective value in order to convert the fraction's denominator to the least common multiple,

\frac{z+1}{x-2y}-\frac{2z-3}{2x-4y}+\frac{z}{3x-6y}

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Simplify,

\frac{z+1}{x-2y}*\frac{6}{6}-\frac{2z-3}{2x-4y}*\frac{3}{3}+\frac{z}{3x-6y}*\frac{2}{2}

\frac{6z+6}{6x-12y}-\frac{6z-9}{6x-12y}+\frac{2z}{6x-12y}

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\frac{6z+6-6z+9+2z}{6x-12y}

\frac{2z+15}{6x-12y}

E)

In this case, one is given the problem that is as follows:

\frac{2}{3d+1}-\frac{1}{5d+3}

Use a similar strategy to solve this problem as used in part (c). Please note that in this case, the least common multiple of the two denominators is the product of the two denominators. In other words, the following value: ((3d+1)(5d+3))

\frac{2}{3d+1}-\frac{1}{5d+3}

\frac{2}{3d+1}*\frac{5d+3}{5d+3}-\frac{1}{5d+3}*\frac{3d+1}{3d+1}

Simplify,

\frac{2}{3d+1}*\frac{5d+3}{5d+3}-\frac{1}{5d+3}*\frac{3d+1}{3d+1}

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\frac{10d+6}{(3d+1)(5d+3)}-\frac{3d+1}{(5d+3)(3d+1)}

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\frac{10d+6-3d-1}{(3d+1)(5d+3)}

\frac{7d+5}{(3d+1)(5d+3)}

\frac{7d+5}{15d^2+14d+3}

F)

The final problem one is given is the following:

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\frac{3a}{2a-3b}-\frac{a+b}{6b-4a}

\frac{3a}{2a-3b}*\frac{-2}{-2}-\frac{a+b}{6b-4a}

Simplify,

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\frac{-6a}{6b-4a}-\frac{a+b}{6b-4a}

\frac{(-6a)-(a+b)}{6b-4a}

\frac{-6a-a-b}{6b-4a}

\frac{-7a-b}{6b-4a}

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