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:
![E(X) = np](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np)
The standard deviation of the binomial distribution is:
![\sqrt{V(X)} = \sqrt{np(1-p)}](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D)
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
,
.
In this problem, we have that:
![p = 0.78, n = 675](https://tex.z-dn.net/?f=p%20%3D%200.78%2C%20n%20%3D%20675)
So
![\mu = E(X) = np = 675*0.78 = 526.5](https://tex.z-dn.net/?f=%5Cmu%20%3D%20E%28X%29%20%3D%20np%20%3D%20675%2A0.78%20%3D%20526.5)
![\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{675*0.78*0.22} = 10.76](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B675%2A0.78%2A0.22%7D%20%3D%2010.76)
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
. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{540 - 526.5}{10.76}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B540%20-%20526.5%7D%7B10.76%7D)
![Z = 1.25](https://tex.z-dn.net/?f=Z%20%3D%201.25)
has a pvalue of 0.8944
89.44% probability that less than 80% of the sample would report eating healthily the previous day