Mathematics, 14.02.2020 00:29 jaqsway34

According to a candy company, packages of a certain candy contain 20% orange candies. Find the approximate probability that the random sample of 400 candies will contain 23% or more orange candies.
Using a normal approximation,
what is the probability that at least 23% of 400 randomly sampled candies will be orange?

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The probability that at least 23% of 400 randomly sampled candies will be orange is 6.68% and this can be determined by using the given data.

Given :

According to a candy company, packages of a certain candy contain 20% orange candies.The random sample of 400 candies will contain 23% or more orange candies.

Given that p = 0.2 and n = 400.

Now, to evaluate the mean the following formula can be used:

$\mu = np = 0.2\times 400$

$\mu = 80$

Now, to evaluate standard deviation following formula can be used:

$\rm \sigma = \sqrt{V(x)}$

$\rm \sigma = \sqrt{np(1-p)}$

$\rm \sigma = \sqrt{400\times 0.2\times 0.8} = 8$

Now, to determine the probability that at least 23% of 400 randomly sampled candies will be orange, first, find the value of X.

$\rm X = 0.23\times 400 = 92$

Now, the value of Z is given by:

$\rm Z = \dfrac{X-\mu }{\sigma}$

$\rm Z=\dfrac{92-80}{8}$

Z = 1.5 has a p-value 0.9332

Now, 1 - 0.9332 = 0.0668.

So, the probability that at least 23% of 400 randomly sampled candies will be orange is 6.68%.

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Answer from: smilingntn33p7pqpp

6.68% probability that at least 23% of 400 randomly sampled candies will be orange.

Step-by-step explanation:

For each candy, there are only two possible outcomes. Either it is orange, or it is not. So we use the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

p = 0.2, n = 400

$\mu = E(X) = np = 400*0.2 = 80$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{400*0.2*0.8} = 8$

Using a normal approximation, what is the probability that at least 23% of 400 randomly sampled candies will be orange?

23% of 400 is 0.23*400 = 92.

So this probability is 1 subtracted by the pvalue of Z when X = 92. So

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

$Z = \frac{92 - 80}{8}$

Z = 1.5

Z = 1.5 has a pvalue of 0.9332.

So there is a 1-0.9332 = 0.0668 = 6.68% probability that at least 23% of 400 randomly sampled candies will be orange.

Answer from: Quest

saving = more money

step-by-step explanation:

if you have 100 bucks, and you want to buy a candy bar, even if you have one at home, if you choose not to buy it, you save.

~courtney

Answer from: Quest

∠cbe and ∠deb are same-side interior angles.

i took the test

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