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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In this problem:
The mean and the standard deviation are given by:
[tex]\mu = np = 300(0.35) = 105[/tex]
[tex]\sigma = \sqrt{np(1-p)} = \sqrt{300(0.35)(0.65)} = 8.26[/tex]
Using continuity correction, the probability that at most 80 of these people will choose the creamery's brand is [tex]P(X \leq 80 + 0.5) = P(X \leq 80.5)[/tex], which is the p-value of Z when X = 80.5, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{80.5 - 105}{8.26}[/tex]
[tex]Z = -2.97[/tex]
[tex]Z = -2.97[/tex] 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