Using the normal distribution, it is found that the probability that the association would observe 654 or less people who drink milk in the sample is of 0%.
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
For the binomial distribution, the parameters are given as follows:
n = 800, p = 0.87.
Hence, the mean and the standard deviation of the approximation are given by:
The probability that the association would observe 654 or less people who drink milk in the sample, using continuity correction, is [tex]P(X \leq 654.5)[/tex], which is the p-value of Z when X = 654.5, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{654.5 - 696}{9.5121}[/tex]
Z = -4.36
Z = -4.36 has a p-value of 0.
0% probability that the association would observe 654 or less people who drink milk in the sample.
More can be learned about the normal distribution at https://brainly.com/question/20909419
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