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

Explanation:

Given : Mean :
`\mu = 15.45`

Standard deviation :
`\sigma = 13.70`

The formula to calculate the z-score :-

`z=(x-\mu)/(\sigma)`

For x= 5 degrees

`z=(5-15.45)/(13.70)=-0.7627737226\approx-0.76`

For x= 10 degrees

`z=(10-15.45)/(13.70)=-0.397810218\approx-0.40`

The P-value :
`P(-0.76<z<-0.40)=P(z<-0.40)-P(z<-0.76)`

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

In percent ,
`0.1210*100=12.10\%`

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