Answer:
P(X < 126) = 0.0418
Step-by-step explanation:
To solve this question, 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 distributions 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].
Sample of 203, 67.5% of the U.S. population were born in their state of residence.
This means that [tex]n = 203, p = 0.675[/tex]
So, for the approximation:
[tex]\mu = E(X) = np = 203*0.675 = 137.025[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{203*0.675*0.325} = 6.67[/tex]
What is the probability that fewer than 126 were born in their state of residence?
Using continuity correction, this is P(X < 126 - 0.5) = P(X < 125.5), which is the pvalue of Z when X = 125.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{125.5 - 137.025}{6.67}[/tex]
[tex]Z = -1.73[/tex]
[tex]Z = -1.73[/tex] has a pvalue of 0.0418
So
P(X < 126) = 0.0418
