Respuesta :
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.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]\mu = np, \sigma = \sqrt{np(1-p)}[/tex].
In this problem, the proportion and the sample size are, respectively, p = 0.22 and n = 200, hence:
[tex]\mu = np = 200(0.22) = 44[/tex]
[tex]\sigma = \sqrt{np(1 - p)} = \sqrt{200(0.22)(0.78)} = 5.8583[/tex]
The probability that more than 30 are single, using continuity correction, is P(X > 30.5), which is 1 subtracted by the p-value of Z when X = 30.5, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{30.5 - 44}{5.8583}[/tex]
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 https://brainly.com/question/24663213
