Normal Distribution Question(Measurement of wind speed)?

Question

Measurements of wind speed on a certain island were taken over a period of one year. A box-and-

whisker plot of the data is obtained where the LQ= 39 and UP=63. It is suggested that wind speed can be modelled approximately by a normal distribution with mean μ and standard deviation σ.

(i) Estimate the value of μ.

(ii) Estimate the value of σ.

The value of μ=(63+39)/2=51, but can someone help me how to find the σ?

Answer 1

Z = (x- mean) / σ

You have two Z value and x value you can use to find σ

LQ = 39 , has P(z< Z) = 0.25

UP= 63, has P(z< Z) = 0.75

Using the lower value

So look up 0.25 inside the z-table

P(z< -0..68) = 0.24825

P(z< -0.67) = 0.25143

Interpolate if you want for a more accurate value (-0.67)

-0.68 + ( 0.25 -0.2485)*( -0.67- (-0.68)) / ( 0.25143- 0.24825)

Z = -0.6753

Now use formula Z = (x- mean) / σ

-0.6753= ( 39-51) / σ = -12/ σ

-0.6753σ= -12

σ= -12/ -0.6753= 17.76988005

====== answer

σ= 17.76988005

==== checking answer

Z = ( 39-51)/ 17.76988005 = -0.6753

Z = (63-51) / 17.76988005 = 0.6753