Let F be an RV that represents the operating temperature in Fahrenheit of one in

Question

Let F be an RV that represents the operating temperature in Fahrenheit of one instance of a manufacturing process, and let F ~ N(100, 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). Using the table provided, solve for the following (it's not a bad idea to double check your answers using R): Find the probability that one randomly selected instance of the process will have operating temperature greater than 98.6 Fahrenheit. Find the distribution of C. Find the probability that one randomly selected instance of the process will have operating temperature below 32 Celsius.

Explanation / Answer

Solution:

We are given that F is normally distributed with mean = 100 and SD = 5

Part a

We have to find P(X>98.6)

P(X>98.6) = 1 – P(X<98.6)

Now, we have to find the Z-score for X = 98.6

The z-score formula is given as below:

Z = (x – mean) / SD

Z = (98.6 – 100) / 5

Z = -0.28

P(X<98.6) = P(Z< -0.28) = 0.389739

P(X>98.6) = 1 – P(X<98.6)

P(X>98.6) = 1 – 0.389739

P(X>98.6) = 0.610261

Required probability = 0.610261

Part b

First we have to convert mean and standard deviation in Celsius. The converting formula is given as below:

C = (5/9)(F – 32)

For mean, C = (5/9)(100 – 32) = 37.78

For Sd, C = (5/9)(5 – 32) = -15 (and/or 15)

[Note: SD cannot be negative value; here this is due to conversion factor and interval scale of the variable.]

Therefore, var = (-15)^2 = 225

This means C follows N(37.78, 225) where mean = 37.78 and variance = 225

Part c

Here, we have to find P(X<32)

Mean = 37.78

SD = 15

Z = (X – mean) / SD

Z = (32 – 37.78) / 15 = -0.38533

P(X<32) = P(Z<-0.38533) = 0.349995

Required probability = 0.349995

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.