Using the normal approximation to the binomial, it is found that the probability is given by:
D. 59.8%
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It 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, 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:
- Males and females are equally probable, hence p = 0.5.
- 65 lambs are born, hence n = 65.
The mean and the standard deviation of the approximation are given by:
[tex]\mu = np = 65(0.5) = 32.5[/tex]
[tex]\sigma = \sqrt{np(1-p)} = \sqrt{65(0.5)(0.5)} = 4.03[/tex]
Using continuity correction, the probability that at least 32 will be female is P(X > 31.5), which is one subtracted by the p-value of Z when X = 31.5, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{31.5 - 32.5}{4.03}[/tex]
[tex]Z = -0.25[/tex]
[tex]Z = -0.25[/tex] has a p-value of 0.402.
1 - 0.402 = 0.598.
Hence option D is correct.
More can be learned about the normal distribution at https://brainly.com/question/24663213
