In December, Seattle's daily high temperature has a mean of 47"F and a standard deviation of 10PF The formula to convert degrees Fahrenheit %F to degrees Celsius %C is: C= ('F -32) Use this information to answer the following questions_ Compute the mean of Seattle's daily high temperature in degrees Celsius C The mean of the daily high temperature in degrees Celcius C (Round your response to three decimal place:) Compute the standard deviation of Seattle's daily high temperature in degrees Celsius %C_ The standard deviation of the daily high temperature in degrees Celisius (Round your response to three decimal places) Compute the variance of Seattle's daily high temperature in degrees Celsius %C_ The variance of the daily high temperature in degrees Celsius = lvc (Round your response to three decimal places)

So for this problem, i'll begin by noting that in general we have that the expected value of- or actually let's see here if we have x, is a random variable. We have a and b are constants. Then we have that the expected value of a x plus b is equal to a times the expected value of x, plus b, and we have that the variance of a x plus b is equal to a squared times. The variance of x so be that other constant, being added on has no part of it, which this also means that the standard deviation of a x plus b will be equal to the square root of a squared. So that's the absolute value of a times. The square root of the variance, so that's times the standard deviation of x, so knowing that we would then have that the mean or that, if let's say we have x, is the temperature in fahrenheit, so we have, the expected value of x is equal to 47. That'S the mean value and the standard deviation of x is equal to 10 point. We have then the expected value of 5 over 9 times x. Minus 32 point: that's going to be equal to 5 over 9 times the expected value of x minus 5 over 9 times 32 point. So once i can here i'll, just calculate that out on screen, so we have 5 over 9 times 47 minus minus 5. Over 9 times 32 point we have that the expected value or the mean value in degrees celsius is going to be 8.33 or 8.333. Then we'll have that the variance is going to be equal to 5 squared or 5 over 9, all squared times our ver r times our standard deviation squared when we put that as a decimal. We have that the variance i know i'm doing this in a slightly different order, are variance, is 30.864, and so our standard deviation will be equal to the square root of 30.864, which then gives a result of 5.556.