Question

the time taken by a student to the university has been shown to be normally distributed with mean of 16 minutes and standard deviation of 2.1 minutes. He walks in once a day during term time, 180 days per year, and leaves home 20 minutes before his first lecture. a. Find the 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.

Answer 1

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 16, \sigma = 2.1$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20 - 16}{2.1}$

$Z = 1.905$

$Z = 1.905$ 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