Question

Daily low temperatures in Columbus, OH in January 2014 were approximately normally distributed with a mean of 15.45 and a standard deviation of 13.70. What percentage of days had a low temperature below 2 degrees? (Enter a number without the percent sign, rounded to the nearest 2 decimal places)

Answer 1

Answer:

16.35%

Step-by-step explanation:

We want probability, x less than 2, which means, we need:

P (x < 2)

First we need to convert x into z-score by using formula:

$z=\frac{x-\mu}{\sigma}$

Where $\mu$ is the mean = 15.45, and

$\sigma$  is the std. deviation = 13.70

Plugging these into the formula, we have:

$z=\frac{x-\mu}{\sigma}\\z=\frac{2-15.45}{13.70}\\z=-0.9817$

So now we have to find:

P (z < -0.9817 ) = 0.1635  [ from z table attached ] [ this is 16.35%]