Answer:
81.85%
Step-by-step explanation:
Given :
The average summer temperature in Anchorage is 69°F.
The daily temperature is normally distributed with a standard deviation of 7°F .
To Find:What percentage of the time would the temperature be between 55°F and 76°F?
Solution:
Mean = \(\mu = 69\)
Standard deviation = \(\sigma = 7\)
Formula : \(z=\frac{x-\mu}{\sigma}\)
Now At x = 55
\(z=\frac{55-69}{7}\)
\(z=-2\)
At x = 76
\(z=\frac{76-69}{7}\)
\(z=1\)
Now to find P(55<z<76)
P(2<z<-1)=P(z<2)-P(z>-1)
Using z table :
P(2<z<-1)=P(z<2)-P(z>-1)=0.9772-0.1587=0.8185
Now percentage of the time would the temperature be between 55°F and 76°F = \(0.8185 \times 100 = 81.85\%\)
Hence If the daily temperature is normally distributed with a standard deviation of 7°F, 81.85% of the time would the temperature be between 55°F and 76°F.
