Answer:
Step-by-step explanation:
Hello!
For any X1, X2,...Xn random variables with binomial distribution, when the sample size is large enough, the distribution of the sample proportion tends to normal with mean np and standard deviation [tex]\sqrt{\frac{p(1-p)}{n} }[/tex].
Symbolically: p[hat]≈N(np;[tex]\frac{p(1-p)}{n}[/tex])
1) For this approximation to be considered valid the following conditions are to be met:
n≥30
n*p≥5
n*(1-p)≥5
2) Since it is an approximation of a normal distribution, the distribution of the sample proportion has a Gaussian form.
3) To calculate probabilities using the approximation to normal, you have to standardize the variable. The statistic is constructed as :
The statistic is [tex]Z= \frac{p[hat] - p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]≈N(0;1)
4)
You have to keep in mind that you are studying proportions, so the interpretation of the results will always be in terms of the population proportion.
For example:
The proportion of brown eggs laid by hybrid hens ( Rhode Island Red and White Star breed mix) is 0.48. If in a production establishment 1200 eggs are selected, what is the probability that at most 45% of the eggs are brown?
In this example, the study variable is X: number of brown eggs in a sample of 1200.
Tha variable has a binomial distribution X~Bi (n;p)
The proportion of brown eggs is 0.48.
The conditions necessary to approximate the distribution of the sample proportion to normal are:
n≥30, in the example n=1200
n*p≥5, in the example n*p= 1200*0.48= 576
n*(1-p)≥5, in the example n*(1-p)= 1200*0.52= 624
All conditions are met, so we can asume that p[hat]≈N(np;[tex]\frac{p(1-p)}{n}[/tex])
And we can use the standard normal approximation to calculate the asked probability:
P(X≤0.45) = P(Z≤[tex]\frac{0.45-0.48}{\sqrt{\frac{0.48*0.52}{1200} } }[/tex])
P(Z≤-2.08) = 0.01876
The probability that the proportion of brown eggs is at most 45% is 0.01876.
I hope it helps!
