Question

. Let F be an RV that represents the operating temperature in Fahrenheit of one instance of a manufacturing process, and let F ∼ N(90, 5 2 ). Let C be an RV that represents the same process, but measured in Celsius. Fahrenheit can be converted to Celsius using C = 5 9 (F − 32). (I recommend doing these with a calculator and N(0, 1) table as practice for the exam. Then check your answers with R if you wish.)

(a) Find the probability that one randomly selected instance of the process will have operating temperature greater than 93.8 Fahrenheit.

(b) C is also normally distributed. Use the properties of E() and V AR() to find its mean and variance.

(c) Find the probability that one randomly selected instance of the process will have operating temperature below 29 Celsius.

(d) Find the Celsius temperature x such that the probability that the operating temperature in Celsius of one instance is less than x is .25.

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	90
std deviation   =σ=	5

probability =P(X>93.8)=P(Z>(93.8-90)/5)=P(Z>0.76)=1-P(Z<0.76)=1-0.7764=0.2236

b)

mean =(5/9)*(90-32)=32.2222

Variance =(5/9)²*Var(F) =(25/81)*25 =7.7160

c)

probability =P(X<29)=(Z<(29-32.2222)/2.778)=P(Z<-1.16)=0.1230

d)

for 25th percentile critical value of z=		-0.67
therefore corresponding value=mean+z*std deviation=32.2222-0.67*2.778 =			30.3611