Using the normal approximation to the binomial, it is found that the probabilities are given as follows:
a) 0.0055 = 0.55%.
b) 0.2296 = 22.96%.
<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
.
The parameters of the binomial distribution are given as follows:
n = 150, p = 0.889.
Hence the mean and the standard deviation of the approximation are:
.
Item a:
Using continuity correction, the probability is P(123.5 < X < 124.5), which is the <u>p-value of Z when X = 124.5 subtracted by the p-value of Z when X = 123.5</u>, hence:
X = 124.5:
Z = -2.3
Z = -2.3 has a p-value of 0.0107.
X = 123.5:
Z = -2.56
Z = -2.56 has a p-value of 0.0052.
Hence the probability is 0.0107 - 0.0052 = 0.0055 = 0.55%.
Item b:
The probability is P(112.5 < X < 130.5), which is the <u>p-value of Z when X = 130.5 subtracted by the p-value of Z when X = 112.5</u>, hence:
X = 130.5:
Z = -0.74
Z = -0.74 has a p-value of 0.2296.
X = 112.5:
Z = -5.42
Z = -5.42 has a p-value of 0.
Hence the probability is 0.2296 - 0 = 0.2296 = 22.96%.
More can be learned about the normal approximation to the binomial at brainly.com/question/14424710
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