Mathematics, 02.07.2021 23:00 lucyamine0

Suppose a large shipment of televisions contained 9% defectives. If a sample of size 393 is selected, what is the probability that the sample proportion will differ from the population proportion by less than 3%

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Answer from: Svetakotok

0.9624 = 96.24% probability that the sample proportion will differ from the population proportion by less than 3%

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Suppose a large shipment of televisions contained 9% defectives

This means that p = 0.09

Sample of size 393

This means that n = 393

Mean and standard deviation:

$\mu = p = 0.09$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.09*0.91}{393}} = 0.0144$

What is the probability that the sample proportion will differ from the population proportion by less than 3%?

Proportion between 0.09 - 0.03 = 0.06 and 0.09 + 0.03 = 0.12, which is the p-value of Z when X = 0.12 subtracted by the p-value of Z when X = 0.06.

X = 0.12

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.12 - 0.09}{0.0144}$

Z = 2.08

Z = 2.08 has a p-value of 0.9812

X = 0.06

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.06 - 0.09}{0.0144}$

Z = -2.08

Z = -2.08 has a p-value of 0.0188

0.9812 - 0.0188 = 0.9624

0.9624 = 96.24% probability that the sample proportion will differ from the population proportion by less than 3%

Answer from: Quest

i can not answer the question due to the fact that you spelled "carton" as "cartoon", a cartoon is something you watch on tv, a carton is a box that holds things. and also you've been waiting for this question for a while so why don't you get off your lazy a** and try to figure it out for yourself for once!

step-by-step explanation:

===============================================================^

look above, and leave your comments down below, and don't for get a five stars and a and brainliest xd

sorry for the stars in the middle of words its because it wouldn't let me submit. xd

How many cartoon of food an how many cartoons of clothing should be sent with each plane? including

Answer from: Quest

vertical asymptote $x=3$ .

horizontal asymptote $y=-1$ .

step-by-step explanation:

given function is $f\left(x\right)=\frac{-3x-5}{3x-9}$ .

vertical asymptote is given by setting

denominator =0

$3x-9=0$

$3x=9$

$x=\frac{9}{3}$

$x=3$

when degree of numerator and denominator are same then divide first term of numerator by first term of denominator to get horizontal asymptote.

$y=\frac{-3x}{3x}$

$y=-1$

Graph all vertical and horizontal asymptotes of the function. f(x)= -3x-5 3x-9 if anyone can it’d

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