Respuesta :
Answer:
89.44% probability that less than 80% of the sample would report eating healthily the previous day
Step-by-step explanation:
We use the binomial approximation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]p = 0.78, n = 675[/tex]
So
[tex]\mu = E(X) = np = 675*0.78 = 526.5[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{675*0.78*0.22} = 10.76[/tex]
What is the approximate probability that less than 80% of the sample would report eating healthily the previous day?
This is the pvalue of Z when [tex]X = 0.8*675 = 540[/tex]. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{540 - 526.5}{10.76}[/tex]
[tex]Z = 1.25[/tex]
[tex]Z = 1.25[/tex] has a pvalue of 0.8944
89.44% probability that less than 80% of the sample would report eating healthily the previous day
