Question

The average daily high temperature in June in LA is 77 degree F with a standard deviation of 5 degree F. Suppose that the temperatures in June closely follow a normal distribution. What is the probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June? How cold are the coldest 10% of the days during June in LA?

Answer 1

Answer:

11.51% probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June

The coldest 10% of the days during June in LA have high temperatures of 70.6F or lower.

Step-by-step explanation:

Problems of normally distributed samples are solved using 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 problem, we have that:

$\mu = 77, \sigma = 5$

What is the probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June?

This probability is 1 subtracted by the pvalue of Z when X = 83. So

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

$Z = \frac{83 - 77}{5}$

$Z = 1.2$

$Z = 1.2$ has a pvalue of 0.8849.

1 - 0.8849 = 0.1151

11.51% probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June

How cold are the coldest 10% of the days during June in LA?

High temperatures of X or lower, in which X is found when Z has a pvalue of 0.1, so whn Z = -1.28

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

$-1.28 = \frac{X - 77}{5}$

$X - 77 = -1.28*5$

$X = 70.6$

The coldest 10% of the days during June in LA have high temperatures of 70.6F or lower.