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3241004551 [841]
4 years ago
12

Determine whether the normal sampling distribution can be used. The claim is p < 0.015 and the sample size is n

Mathematics
1 answer:
Svetradugi [14.3K]4 years ago
8 0

Complete Question

Determine whether the normal sampling distribution can be used. The claim is p < 0.015 and the sample size is n=150

Answer:

Normal sampling distribution can not be used

Step-by-step explanation:

From the question we are told that

    The  null hypothesis is  H_o  :  p =  0.015

     The  alternative hypothesis is    H_a  :  p  <  0.015

     

The  sample size is  n= 150

Generally in order to use  normal sampling distribution  

     The value  np  \ge  5

So  

         np =  0.015 * 150

         np =  2.25

Given that  np < 5   normal sampling distribution  can not be used

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erma4kov [3.2K]
<h3>Answer:</h3>

\huge{\boxed{n=-4}}

<h3>Explanation:</h3>

\begin{array}{cc}\begin{flushright}\text{Divide both sides of the equation by 3.}\end{flushright}&\begin{flushleft}-4n-9=7\end{flushleft}\\\begin{flushright}\text{Add 9 on both sides.}\end{flushright}&\begin{flushleft}-4n=16\end{flushleft}\\\begin{flushright}\text{Divide both sides by -4.}\end{flushrigh}&n=-4\end{array}

7 0
4 years ago
Read 2 more answers
The weight of people in a small town in Missouri is known to be normally distributed with a mean of 186 pounds and a standard de
OleMash [197]

Answer:

the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds is 0.0166

Step-by-step explanation:

The summary of the given statistical data set are:

Sample Mean = 186

Standard deviation = 29

Maximum capacity 3,417 pounds or 17 persons.

sample size = 17

population mean =3417

The objective is to determine the probability  that a random sample of 17 persons will exceed the weight limit of 3,417 pounds

In order to do that;

Let assume X to be the random variable that follows the normal distribution;

where;

Mean \mu = 186 × 17 = 3162

Standard deviation = 29* \sqrt{17}

Standard deviation = 119.57

P(X>3417) = P(\dfrac{X - \mu}{\sigma}>\dfrac{X - \mu}{\sigma})

P(X>3417) = P(\dfrac{3417 - \mu}{\sigma}>\dfrac{3417 - 3162}{119.57})

P(X>3417) = P(Z>\dfrac{255}{119.57})

P(X>3417) = P(Z>2.133)

P(X>3417) =1- 0.9834

P(X>3417) =0.0166

Therefore; the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds is 0.0166

5 0
3 years ago
What algebraic expression is, a to the 3rd power, increased by the product of 5 and c?
JulijaS [17]

Step-by-step explanation:

{a}^{3}  + 5c

5 0
3 years ago
MAX POINTS!
zheka24 [161]

<u>Answer:</u>

\frac{200}{3} square units

<u>Step-by-step explanation:</u>

Knowing the function and that the triangle is formed by both the x and y-axis, you already have the triangle's height, which is 10 (the y-intercept of the functon) To find the base length, all you have to do is find the x-intercept. To do that, simply:

Set function equal to 0 and solve for x:

f(x) = 10 - \frac{3}{4} x

0 = 10 - \frac{3}{4} x

\frac{3}{4} x = 10

x = \frac{40}{3}

This means that the base length of the triangle is also  \frac{40}{3}. Now that you have the height and base length, you can

Plug values into the formula for the area of a triangle:

The area of a triangle is \frac{1}{2} × height × base

Height = 10, base = \frac{40}{3}

Area = \frac{1}{2} × 10 × \frac{40}{3}

Area = 5 × \frac{40}{3}

Area =  \frac{200}{3}

8 0
3 years ago
If F(X) = -4X + 3, what is F(-2)
Setler79 [48]

Answer:

11

Step-by-step explanation:

When you plug in -2 for x you get -4(-2) +3

8 +3 = 11

F(-2)=11

Hope this helps, have a nice day

7 0
3 years ago
Read 2 more answers
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