Respuesta :
Using the normal distribution, there is a 0.7967 = 79.67% probability that at most 12 adults in a sample of 100 adults pass this fitness test.
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].
The parameters of the binomial distribution are given by:
p = 0.1, n = 100.
Hence the mean and the standard deviation for the approximation are given by:
- [tex]\mu = np = 100 \times 0.1 = 10[/tex]
- [tex]\sigma = \sqrt{np(1-p)} = \sqrt{100 \times 0.1 \times 0.9} = 3[/tex]
The probability at most 12 adults in a sample of 100 adults pass this fitness test, using continuity correction, is P(X < 12.5), which is the p-value of Z when X = 12.5, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{12.5 - 10}{3}[/tex]
Z = 0.83
Z = 0.83 has a p-value of 0.7967.
0.7967 = 79.67% probability that at most 12 adults in a sample of 100 adults pass this fitness test.
More can be learned about the normal distribution at https://brainly.com/question/28159597
#SPJ1
