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 between 5 degrees and 10 degrees? (Enter a number without the percent sign, rounded to the nearest 2 decimal places)

Answer 1

Answer: 12.10

Step-by-step explanation:

Given : Mean : $\mu = 15.45$

Standard deviation : $\sigma = 13.70$

The formula to calculate the z-score :-

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

For x= 5 degrees

$z=\dfrac{5-15.45}{13.70}=-0.7627737226\approx-0.76$

For x= 10 degrees

$z=\dfrac{10-15.45}{13.70}=-0.397810218\approx-0.40$

The P-value : $P(-0.76$

$=0.3445783-0.2236273=0.120951\approx0.1210$

In percent , $0.1210\times100=12.10\%$

Hence, the percentage of days had a low temperature between 5 degrees and 10 degrees = 12.10%