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

16.35%

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=(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=(x-\mu)/(\sigma)\\z=(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%]