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The main bearing clearance (in mm) in a certain type of
Chapter 2, Problem 20E(choose chapter or problem)
The main bearing clearance (in mm) in a certain type of engine is a random variable with probability density function
\(f(x)=\left\{\begin{array}{cl}
625 x & 0<x \leq 0.04 \\
50-625 x & 0.04<x \leq 0.08 \\
0 & \text { otherwise }
\end{array}\right.
\)
a. What is the probability that the clearance is less than 0.02 mm?
b. Find the mean clearance.
c. Find the standard deviation of the clearances.
d. Find the cumulative distribution function of the clearance.
e. Find the median clearance.
f. The specification for the clearance is 0.015 to 0.063 mm. What is the probability that the specification is met?
Equation Transcription:
Text Transcription:
f(x)=625 0<x0.04
f(x)=50-625x 0.04<x0.08
f(x)=0 otherwise
Questions & Answers
QUESTION:
The main bearing clearance (in mm) in a certain type of engine is a random variable with probability density function
\(f(x)=\left\{\begin{array}{cl}
625 x & 0<x \leq 0.04 \\
50-625 x & 0.04<x \leq 0.08 \\
0 & \text { otherwise }
\end{array}\right.
\)
a. What is the probability that the clearance is less than 0.02 mm?
b. Find the mean clearance.
c. Find the standard deviation of the clearances.
d. Find the cumulative distribution function of the clearance.
e. Find the median clearance.
f. The specification for the clearance is 0.015 to 0.063 mm. What is the probability that the specification is met?
Equation Transcription:
Text Transcription:
f(x)=625 0<x0.04
f(x)=50-625x 0.04<x0.08
f(x)=0 otherwise
ANSWER:
Solution
Step 1 of 6 :
The main bearing clearance in a in a certain type of engine is a random variable with the probability density function is
Our goal is to find :
a). What is the probability that the clearance is less than 0.02 mm?
b). Find the mean clearance.
c). Find the standard deviation of the clearance.
d). Find the cumulative distribution function of the clearance.
e). Find the median clearance.
f). Find the probability that the specification is met.
a).
Now we have to find the probability that the clearance is less than 0.02 mm.
So P(X < 0.02) is
The probability density function is
Here we take the limits 0 to 0.02.
Then,
P(X < 0.02) =
We are integrating.
P(X < 0.02) =
P(X < 0.02) =
P(X < 0.02) =
P(X < 0.02) =
P(X < 0.02) =
Therefore the probability that the clearance is less than 0.02 mm is 0.125.