According to a candy​ company, packages of a certain candy contain 1616​% orange candies. Suppose we examine 100100 random candies. a. What value should we expect for our sample percentage of orange​ candies? b. What is the standard​ error? c. Use your answers to fill in the blanks below. We expect​ ____% orange​ candies, give or take​ _____%.

Answer :

Answer:

a) By the Central Limit Theorem, 16%.

b) 0.0367 = 3.67%

c) We expect 16% orange​ candies, give or take 3.67%.

Step-by-step explanation:

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

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

For samples of size n of a proportion p, the expected sample percentage is p and the standard error is [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

In this problem, we have that:

[tex]p = 0.16, n = 100[/tex]

a. What value should we expect for our sample percentage of orange​ candies?

By the Central Limit Theorem, 16%.

b. What is the standard​ error?

[tex]s = \sqrt{\frac{0.16*0.84}{100}} = 0.0367[/tex]

0.0367 = 3.67%

c. Use your answers to fill in the blanks below. We expect​ ____% orange​ candies, give or take​ _____%.

We expect 16% orange​ candies, give or take 3.67%.

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