Question

Stattles candies have five different colors that are equally distributed (20% orange), and each bag represents a
random sample of the entire population of Stattles. A 14 ounce bag of Stattles has 220 of the multi-colored
candies.
a. Cindy loves orange Stattles and is interested in exploring how many orange Stattles she can expect to get in
a bag. Demonstrate that the sampling distribution of p, the proportion of orange Stattles in a bag, satisfies all
three conditions of inference.
b. Find the mean and standard deviation of the sampling distribution of p.
c. Sketch and label the sampling distribution curve using your values from part b.
d. Calculate the probability that Cindy will get at least 50 orange Stattles in her next 14 ounce bag.

Answer 1

Using the normal approximation to the binomial, we have that:

a)

Since $np \geq 10$ and $n(1 - p) \geq 10$, and the candies are independent, that is, they have the same probability of being orange, the conditions of inference are satisfied.

b) The mean is of 44 and the standard deviation is of 5.93.

c) The sketch is given at the end of this answer.

d) 0.1762 = 17.62% probability that Cindy will get at least 50 orange Stattles in her next 14 ounce bag.

Binomial probability distribution

Probability of x successes on n trials, with p probability.

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 distribution:

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

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

• It 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, which is the percentile of X.

• If both $np \geq 10$ and $n(1 - p) \geq 10$, the binomial distribution can be approximated to the normal.

In this problem:

• 220 candies, thus $n = 220$
• 20% are orange, thus $p = 0.2$

Item a:

The conditions are:

$np = 220(0.2) = 44 \geq 10$

$n(1 - p) = 220(0.8) = 176 \geq 10$

Since $np \geq 10$ and $n(1 - p) \geq 10$, and the candies are independent, that is, they have the same probability of being orange, the conditions of inference are satisfied.

Item b:

The mean is:

$E(X) = np = 220(0.2) = 44$

The standard deviation is:

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{220(0.2)(0.8)} = 5.93$

The mean is of 44 and the standard deviation is of 5.93.

Item c:

Considering the mean and standard deviation found in item b, the sketch of the normal distribution is given at the end of this answer.

Item d:

Using continuity correction, this probability is $P(X \geq 50 - 0.5) = P(X \geq 49.5)$, which is 1 subtracted by the p-value of Z when X = 49.5. Then:

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

$Z = \frac{49.5 - 44}{5.93}$

$Z = 0.93$

$Z = 0.93$ has a p-value of 0.8238.

1 - 0.8238 = 0.1762

0.1762 = 17.62% probability that Cindy will get at least 50 orange Stattles in her next 14 ounce bag.

A similar problem is given at https://brainly.com/question/24261244