Respuesta :
Using the normal distribution, there is a 0.7123 = 71.23% probability that at least 45 of them have a football team.
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 for the binomial distribution are given as follows:
n = 70, p = 534/800 = 0.6675.
Hence the mean and the standard deviation of the normal approximation are given by:
- [tex]\mu = np = 70 \times 0.6675 = 46.725[/tex]
- [tex]\sigma = \sqrt{np(1 - p)} = \sqrt{70 \times 0.6675 \times 0.3325} = 3.94[/tex]
Using continuity correction, the probability that at least 45 of them have a football team is one subtracted by the p-value of Z when X = 44.5, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (44.5 - 46.725)/3.94
Z = -0.56
Z = -0.56 has a p-value of 0.2877.
1 - 0.2877 = 0.7123.
0.7123 = 71.23% probability that at least 45 of them have a football team.
More can be learned about the normal distribution at https://brainly.com/question/4079902
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