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3 years ago

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Suppose an automotive repair company wants to determine the current percentage of customers who keep up with regular vehicle mai

ntenance. How many customers should the company survey in order to be 90% confident that the estimated (sample) proportion is within 2 percentage points of the true population proportion of customers who keep up with regular vehicle maintenance?

2 answers:

Nina [5.8K]3 years ago

8 0

Answer: 1692

Step-by-step explanation:

Formula to find the sample size :

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

Given : Confidence level : $(1-\alpha)=0.90$

⇒ significance level = $\alpha= 0.10$

z-value for 90% confidence interval (using z-table)= $z_{\alpha/2}=1.645$

Prior estimate of the population proportion (p) of customers who keep up with regular vehicle maintenance is unknown.

Let we take p= 0.5

Margin of error : E= 2%=0.02

Now, the required sample size will be :

$n=0.5(1-0.5)(\dfrac{1.645}{0.02})^2$

Simplify , we get

$n=(0.25)(6765.0625)=1691.265625\approx1692$

Hence, the required sample size = 1692

Zielflug [23.3K]3 years ago

7 0

Answer: 2401

Step-by-step explanation:

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3 years ago

Let R(x), C(x), and P(x) be, respectively, the revenue, cost, and profit, in dollars, from the production and sale of x it

sp2606 [1]

Answer:

$P(x) = -0.004x^2 - 3.7x - 50$

Step-by-step explanation:

We have to find profit $P(x)$ .

This can be easily found using a formula for Profit given the Revenue and Cost.

$Profit = Revenue - Cost$

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$R(x) = 6x$

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to find P(x) we can simply subtract R(x) by C(x).

$P(x) = R(x) - C(x)$

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and finally, after simplify this equation subtracting 6x by 2.3x.

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3 0

3 years ago

PLEASE HELP!!!!!!!!!!

kobusy [5.1K]

73 feet per second

Steps-
Miles to cover- 50 miles = 50*5280 feet = 264,000 feet
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2 years ago

A marketing firm is considering making up to three new hires. Given its specific needs, the management feels that there is a 50%

IRISSAK [1]

Answer:

a) 0.9

b) Mean = 1.58

Standard Deviation = 0.89

Step-by-step explanation:

We are given the following in the question:

A marketing firm is considering making up to three new hires.

Let X be the variable describing the number of hiring in the company.

Thus, x can take values 0,1 ,2 and 3.

$P(x\geq 2) = 50\%= 0.5\\P(x = 0) = 10\% = 0.1\\P(x = 3) = 18\% = 0.18$

a) P(firm will make at least one hire)

$P(x\geq 2) = P(x=2) + P(x=3)\\0.5 = P(x=2) + 0.18\\ P(x=2) = 0.32$

Also,

$P(x= 0) +P(x= 1) + P(x= 2) + P(x= 3) = 1\\ 0.1 + P(x= 1) + 0.32 + 0.18 = 1\\ P(x= 1) = 1- (0.1+0.32+0.18) = 0.4$

$\text{P(firm will make at least one hire)}\\= P(x\geq 1)\\=P(x=1) + P(x=2) + P(x=3)\\ = 0.4 + 0.32 + 0.18 = 0.9$

b) expected value and the standard deviation of the number of hires.

$E(X) = \displaystyle\sum x_iP(x_i)\\=0(0.1) + 1(0.4) + 2(0.32)+3(0.18) = 1.58$

$E(x^2) = \displaystyle\sum x_i^2P(x_i)\\=0(0.1) + 1(0.4) + 4(0.32) +9(0.18) = 3.3\\V(x) = E(x^2)-[E(x)]^2 = 3.3-(1.58)^2 = 0.80\\\text{Standard Deviation} = \sqrt{V(x)} = \sqrt{0.8036} = 0.89$

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3 years ago