Question

Data accumulated by the National Climatic Data Center shows that the average wind speed in miles per hour for St. Louis,Missouri, is 9.7. Suppose wind speed measurements are normally distributed for a given geographic location. If 22.45% of the time the wind speed measurements are more than 11.6 miles per hour, what is the standard deviation of wind speed in St. Louis?

Answer 1

Answer:

The  value is  $\sigma = 8.5$    

Step-by-step explanation:

From the question we are told that

  The  population mean is  $\mu = 9.7$

   The  proportion is  $P(X >x) = 0.2245$

   The value considered is  $x = 11.6$

Generally given that the speed measurement is normally distributed we have that

   $P(X > 11.6) = P(\frac{ X - \mu }{ \sigma } > \frac{11.6 - 9.7}{ \sigma } ) = 0.2245$

Generally

   $\frac{X - \mu}{ \sigma } = Z(The \ z -score \ of \ X )$

$P(X > 11.6) = P(Z> \frac{11.6 - 9.7}{ \sigma } ) = 0.2245$

Hence

     $\frac{ 11.6 - 9.7}{ \sigma} = 0.2245$

  $\sigma = 8.5$