If your mom has a big p3n1s, but your dad also has a p3n1s... then how were you born?
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Using the normal approximation to the binomial, it is found that there is a 0.0107 = 1.07% probability that more than 30 are single.
<h3>Normal Probability Distribution</h3>The z-score of a measure X of a normally distributed variable with mean and standard deviation
is given by:
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
- The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with
.
In this problem, the proportion and the sample size are, respectively, p = 0.22 and n = 200, hence:
The probability that more than 30 are single, using continuity correction, is P(X > 30.5), which is <u>1 subtracted by the p-value of Z when X = 30.5</u>, hence:
Z = -2.3
Z = -2.3 has a p-value of 0.0107.
0.0107 = 1.07% probability that more than 30 are single.
More can be learned about the normal distribution at brainly.com/question/24663213