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Anni [7]
3 years ago
7

Select the correct answer. A company offers its employees free recreational activities and gym memberships. Of the 200 employees

, 95 employees registered for recreational activities, 80 employees registered for gym memberships, and 30 employees did not register at all. If an employee is chosen at random, what is the probability that the employee has registered for both recreational activities AND gym memberships?
Mathematics
2 answers:
topjm [15]3 years ago
5 0

By using Venn Diagram, one can easily visualize that 5 people out of 200 chose both the activities(or whatever)

So.

\frac{5}{200}  =  \boxed{\frac{1}{40}}  = 2.5percent

(sorry for the<u> s t u p i d</u> handwriting)

Sophie [7]3 years ago
4 0

Answer:

1/40

Step-by-step explanation:

95 + 80 + 30 = 205
205 - 200 = 5
5/200 = 1/40

Took the test and got it right.

Hope this helped!

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Answer:

C. n=423

Step-by-step explanation:

1) Notation and important concepts

Margin of error for a proportion is defined as "percentage points your results will differ from the real population value"

Confidence=90%=0.9

\alpha=1-0.9=0.1 represent the significance level defined as "a measure of the strength of the evidence that must be present in your sample before you will reject the null hypothesis and conclude that the effect is statistically significant".

\hat p=0.5 represent the sample proportion of consumers in the Oconee County area who would react favorably to a marketing campaig. For this case since we don't have enough info we use the value of 0.5 since is equiprobable the event analyzed.

z_{\alpha/2} represent a quantile of the normal standard distribution that accumulates {\alpha/2} on each tail of the distribution.

2) Formulas and solution for the problem

For this case the margin of error for a proportion is given by this formula

ME=z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}   (1)

For this case the confidence level is 90% or 0.9 so then the significance would be

\alpha=1-0.9=0.1 and \alpha/2=0.05

With \alpha/2=0.05, we can find the value for z_{\alpha/2} using the normal standard distribution table or excel.

The calculated value is z_{\alpha/2}=1.644854

Now from equation (1) we need to solve for n in order to answer the question.

\frac{ME}{z_{\alpha/2}}=\sqrt{\frac{\hat p(1-\hat p)}{n}}  

Squaring both sides:

(\frac{ME}{z_{\alpha/2}})^2=\frac{\hat p(1-\hat p)}{n}

And solving for n we got:

n=\frac{\hat p(1-\hat p)}{(\frac{ME}{z_{\alpha/2}})^2}

Now we can replpace the values

n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.644854})^2}=422.739

And rounded up to the nearest integer we got:

n=423

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