Question

A recent report found that a local creamery has a 35% market share in a certain region. Employees in the marketing
department of the creamery conduct a survey to confirm the results by randomly selecting 300 residents from the regon
The employees asked the question which brand of ice cream do you usually purchase to each resident Use a normal
approximation to calculate the probability that at most 80 of these people will choose the creamery's brand. Use a T1-83,
83 plus, or T-34 calculator to find the probability
- Round your answer to four decimal places

Answer 1

Using the normal approximation to the binomial, it is found that there is a 0.0015 = 0.15% probability that at most 80 of these people will choose the creamery's brand.

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.

• The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$.

In this problem:

• 35% market share, hence $p = 0.35$.
• 300 residents are surveyed, hence $n = 300$.

The mean and the standard deviation are given by:

$\mu = np = 300(0.35) = 105$

$\sigma = \sqrt{np(1-p)} = \sqrt{300(0.35)(0.65)} = 8.26$

Using continuity correction, the probability that at most 80 of these people will choose the creamery's brand is $P(X \leq 80 + 0.5) = P(X \leq 80.5)$, which is the p-value of Z when X = 80.5, thus:

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

$Z = \frac{80.5 - 105}{8.26}$

$Z = -2.97$

$Z = -2.97$ has a p-value of 0.0015.

0.0015 = 0.15% probability that at most 80 of these people will choose the creamery's brand.

A similar problem is given at https://brainly.com/question/25298431