A student would like to estimate the proportion of students at his school who have lunch during period 4. To do so,
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Using the normal distribution, it is found that 7.64% of of sample means are greater than 8.8 hours.
<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.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation
.
The parameters are given as follows:
The proportion of sample means greater than 8.8 hours is <u>one subtracted by the p-value of Z when X = 8.8</u>, hence:
By the Central Limit Theorem
Z = 1.43
Z = 1.43 has a p-value of 0.9236.
1 - 0.9236 = 0.0764.
7.64% of of sample means are greater than 8.8 hours.
More can be learned about the normal distribution at brainly.com/question/25800303
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