Calculate the mean and standard deviation. If the temperatures had been measured

Question

Calculate the mean and standard deviation. If the temperatures had been measured in degrees centigrade, what would the mean and standard deviation be? [Hint: if F and C represent a given temperature in Fahrenheit and Centigrade, then C = (F - 32) times 5/9] The altitudes of four mountain states of about the same size are approximately as follows (feet above sea level): Arizona 4100 Nevada 5500 Colorado 6900 Wyoming 6700 To calculate the mean and standard deviation, it would be natural to just drop the last two Zeros. Would this give you the right answer? Why or why not? Then calculate the mean altitude and standard deviation. What are the mean and standard deviation in yards above sea level? An agricultural experimental station has five square plots, whose This theorem is proved in Appendix 2-5, and may be interpreted very simply: If the individual observations are linearly transformed, then the mean observation is transformed in exactly the same way, and the standard deviation is changed by the factor |b|, with no effect from a. The temperature inside an experimental solar heater was measured in four different spots (degrees Fahrenheit): 238, 227, 220, 235. Calculate the mean and standard deviation. If the temperatures had been measured in degrees centigrade, what would the mean and standard deviation be? [Hint: If F and C represent a given temperature in Fahrenheit and Centigrade, then C = (F - 32) x 5/9] The altitudes of four mountain states of about the same size are approximately as follows (feet above sea level):

Explanation / Answer

degrees F : 238, 227, 220, 235

mean=(µ) = SX/N= (sum of all data)/(number of data) = 920/4=230

Standard deviation = s = sqrt [(S((X-µ)^2))/(N)].
=sqrt[((238-230)2 +(227-230)2 +(220-230)2 +(235-230)2)/4]
= sqrt[198/4]=sqrt[49.5]= 7.035624

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with C=(F-32)*(5/9), the data is:{ 114.4444444 108.3333333 104.4444444 112.7777778}

Following same procedures from before,

µ=110
s=3.90868

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