According to a candy company, packages of a certain

Ответ:

makennahudson94

18.01.2022

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%.

For more information, refer to the link given below:

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Ответ:

lberman2005p77lfi

18.01.2022

Mathematics

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .