All categories

4 years ago

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 $np \ge 5$

$np = 0.015 * 150$

$np = 2.25$

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

You might be interested in

3(-4n - 9) = 21 plc solve

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