Mathematics, 29.06.2021 20:20 182075

Suppose a large consignment of televisions contained 11% defectives. If a sample of size 237 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: solikhalifeoy3j1r

0.8612 = 86.12% 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 consignment of televisions contained 11% defectives.

This means that p = 0.11

Sample of size 237

This means that n = 237

Mean and standard deviation:

$\mu = p = 0.11$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.11*0.89}{237}} = 0.0203$

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

P-value of Z when X = 0.11 + 0.03 = 0.14 subtracted by the p-value of Z when X = 0.11 - 0.03 = 0.08. So

X = 0.14

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

By the Central Limit Theorem

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

$Z = \frac{0.14 - 0.11}{0.0203}$

Z = 1.48

Z = 1.48 has a p-value of 0.9306

X = 0.08

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

$Z = \frac{0.08 - 0.11}{0.0203}$

Z = -1.48

Z = -1.48 has a p-value of 0.0694

0.9306 - 0.0694 = 0.8612

0.8612 = 86.12% probability that the sample proportion will differ from the population proportion by less than 3%

Answer from: Quest

answer: 35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %, p(e) = 27 % and $p(l\cap e)$ = 4 %

and, we have to find out, $p(l\cup e)$

but, we know that,

$p(l\cup e)= p(l) + p(e) - p(l\cap e)$

which is the required equation that will use to find the probability of a late departure or an early departure.

$p(l\cup e)= 12 + 27 - 4$

$p(l\cup e)= 35 %$

therefore, the probability of a late departure or an early arrival = 35 %

Answer from: Quest

a reflection across the x axis and a 90 degree rotation left and right.

Easy easy 4th grade math for 20 points waaaaait 20 yes 20 ! and only 1 question answer

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Suppose a large consignment of televisions contained 11% defectives. If a sample of size 237 is sele...

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