91Ó°ÊÓ

At a certain location the temperature in degrees Fahrenheit at a certain location is normally distributed with mean 68 degrees and standard deviation is 4 degrees.

If X denotes the temperature in degrees Fahrenheit, Y denotes the temperature in degrees Celsius.

Then,\(Y = \frac{{5\left( {X - 32} \right)}}{9}\)

Y is the linear function of X, then Y will also have a normal distribution.

Also,

Mean of the distribution:

\(\begin{array}{c}E\left( Y \right) = \frac{5}{9}\left( {68 - 32} \right)\\ = \frac{5}{9}\left( {36} \right)\end{array}\)

\(\begin{array}{c}E\left( Y \right) = \frac{{180}}{9}\\ = 20\end{array}\)

Variance of the distribution:

\(\begin{array}{c}Var\left( Y \right) = {\left( {\frac{5}{9}} \right)^2}\left( {16} \right)\\ = \frac{{25}}{{81}}\left( {16} \right)\\ = \frac{{400}}{{81}}\\ = 4.938\end{array}\)

Therefore, the distribution of the temperature in degrees Celsius at the same location is the normal distribution with mean 20 and variance 4.938