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Sagot :
Using the normal distribution, it is found that there is a 0.0005 = 0.05% probability of getting more than 66 heads.
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].
For the binomial distribution, the parameters are given as follows:
n = 100, p = 0.5.
Hence the mean and the standard deviation of the approximation are given as follows:
- [tex]\mu = np = 100(0.5) = 50[/tex].
- [tex]\sigma = \sqrt{np(1-p)} = \sqrt{100(0.5)(0.5)} = 5[/tex]
Using continuity correction, the probability of getting more than 66 heads is P(X > 66 + 0.5) = P(X > 66.5), which is one subtracted by the p-value of Z when X = 66.5.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{66.5 - 50}{5}[/tex]
Z = 3.3
Z = 3.3 has a p-value of 0.9995.
1 - 0.9995 = 0.0005.
0.0005 = 0.05%
More can be learned about the normal distribution at https://brainly.com/question/4079902
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