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Fiesta28 [93]
3 years ago
15

For a population with a mean of 40 and a standard deviation of 8 find the z-score corresponding to each of the following samples

.
X = 34 for a sample of n = 1 score:

M = 34 for a sample of n = 4

M = 34 for a sample of n = 16
Mathematics
1 answer:
aleksley [76]3 years ago
3 0

Answer:

a) -0.75

b) -1.5

c) -3

Step-by-step explanation:

The question asks to convert a given value to z score with different sample sizes.

<u>We are given:</u>

Population Mean = u = 40

and

Standard deviation = s = 8

The task is to convert 34 to z-score using 3 different sample sizes.

For a given sample size "n", the formula to calculate the z-score is:

z=\frac{x-u}{\frac{s}{\sqrt{n} } }

For x = 34 and n=1, we get:

z-score=\frac{34-40}{\frac{8}{\sqrt{1} } } = -0.75

For x = 34 and n = 4, we get:

z-score=\frac{34-40}{\frac{8}{\sqrt{4} } } = -1.5

For x = 34 and n = 16, we get:

z-score=\frac{34-40}{\frac{8}{\sqrt{16} } } = -3

From here we can see that increasing the sample sizes increases the magnitude of z-score.

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Answer:

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Step-by-step explanation:

First let's simplify the equation so it will be easier to PEMDAS the equation later.

-7v+4(2-5v)=-154

-7v+8-20v=-154

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Now you must isolate the v variable. You do this by subtracting 8 first and then divide by -27.

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The derivative of the given function is f'(x) = k f(x) where k= \lim_{h \to 0} \frac{f(h)-1}{h}.

<h3>What is the derivative of a function?</h3>

Let f be a function defined on a neighborhood of a real number a. Then f is said to be differentiable or derivable at 'a' if \lim_{x \to a} \frac{f(x)-f(a)}{x-a} exists finitely. The limit is called the derivative or differential coefficient of f at 'a'. It is denoted by f'(a).

If f is differentiable at 'a', then

f'(a) = \lim_{h\to 0} \frac{f(a+h)-f(a)}{h}

<h3>Calculation:</h3>

The given properties are:

(i) f(x + y) = f(x)f(y) for all real numbers x and y.

(ii) \lim_{h \to 0} \frac{f(h)-1}{h} = k; where k is a nonzero real number.

Then, the derivative of the function f(x) is,

f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

From property (i), f(x + h) = f(x)f(h)

On substituting,

f'(x) = \lim_{h \to 0} \frac{f(x)f(h)-f(x)}{h}

      = \lim_{h \to 0} \frac{f(x)[f(h) - 1]}{h}

From property (ii), \lim_{h \to 0} \frac{f(h)-1}{h} = k;

f'(x) = \lim_{h \to 0} \frac{f(x)[f(h) - 1]}{h}

      = f(x). \lim_{h \to 0} \frac{f(h)-1}{h}

      = f(x). k

      = kf(x)

Therefore, f'(x) = k f(x); where f'(x) exists for all real numbers of x.

Learn more about the derivative of a function here:

brainly.com/question/5313449

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Answer:Option A.

Step-by-step explanation:

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