Suppose that a company claims that its chocolate bars weigh 1.6 ounces on average. It took many large samples, and each time the
2 answers:
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Answer:
a) 2.84% probability that he is late for his first lecture.
b) 5.112 days
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
a. Find the probability that he is late for his first lecture.
This is the probability that he takes more than 20 minutes to walk, which is 1 subtracted by the pvalue of Z when X = 20. So
has a pvalue of 0.9716
1 - 0.9716 = 0.0284
2.84% probability that he is late for his first lecture.
b. Find the number of days per year he is likely to be late for his first lecture.
Each day, 2.84% probability that he is late for his first lecture.
Out of 180
0.0284*180 = 5.112 days