MATH SOLVE

3 months ago

Q:
# 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)

Accepted Solution

A:

Answer: 12.10Step-by-step explanation:Given : Mean : [tex]\mu = 15.45[/tex]Standard deviation : [tex]\sigma = 13.70[/tex]The formula to calculate the z-score :-[tex]z=\dfrac{x-\mu}{\sigma}[/tex]For x= 5 degrees[tex]z=\dfrac{5-15.45}{13.70}=-0.7627737226\approx-0.76[/tex]For x= 10 degrees[tex]z=\dfrac{10-15.45}{13.70}=-0.397810218\approx-0.40[/tex]The P-value : [tex]P(-0.76<z<-0.40)=P(z<-0.40)-P(z<-0.76)[/tex][tex]=0.3445783-0.2236273=0.120951\approx0.1210[/tex]In percent , [tex]0.1210\times100=12.10\%[/tex]Hence, the percentage of days had a low temperature between 5 degrees and 10 degrees = 12.10%