I have no idea how to or how I would even start to write a problem like this, Please help.
1 answer:
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Using the normal distribution, it is found that there is a 0.0005 = 0.05% probability of getting more than 66 heads.
<h3>Normal Probability Distribution</h3>The z-score of a measure X of a normally distributed variable with mean and standard deviation
is given by:
- 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
.
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:
.
Using continuity correction, the probability of getting more than 66 heads is P(X > 66 + 0.5) = P(X > 66.5), which is <u>one subtracted by the p-value of Z when X = 66.5</u>.
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 brainly.com/question/4079902
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