Normal Distribution Question(Measurement of wind speed)?

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?

1 Answer

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  • Alan
    Lv 7
    4 years ago

    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

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