I\'m reviewing for my first midterm and this is an old HW problems that I wasn\'
ID: 3124796 • Letter: I
Question
I'm reviewing for my first midterm and this is an old HW problems that I wasn't able to solve. Please show your work, and be as clear as possible. Thanks!
6. Let F be an RV that represents the operating temperature in Fahrenheit of one instance of a manufacturing process, and let F ~ N(90,52). Let C be an RV that represents the same process, but measured in Celsius. Fahrenheit can be converted to Celsius using C(F-32). Using the table provided, solve for the following ( it's not a bad idea to double check your answers using R ) (a) Find the probability that one randomly selected instance of the process will have operating temperature greater than 93.8 Fahrenheit (b) Find the distribution of C. (Hint: C~?(?,?) (c) Find the probability that one randomly selected instance of the process will have operating temperature below 29 Celsius.Explanation / Answer
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 93.8
u = mean = 90
s = standard deviation = 5
Thus,
z = (x - u) / s = 0.76
Thus, using a table/technology, the right tailed area of this is
P(z > 0.76 ) = 0.223627292 [ANSWER]
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b)
As
C = (5/9)(F-32)
Then
u(C) = 5/9 u(F) - (5/9)(32) = (5/9)*90 - (5/9)*32 = 32.22222222
sigma^2(C) = (5/9)^2 (5^2) = 7.716049383
Thus,
C ~ (32.22222, 7.71604) [ANSWER]
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c)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 29
u = mean = 32.2222222
s = standard deviation = sqrt(7.716049) = 2.777777778
Thus,
z = (x - u) / s = -1.159999992
Thus, using a table/technology, the left tailed area of this is
P(z < -1.159999992 ) = 0.123024405 [ANSWER]