Question
Failures in machine in an 8 hrs shift is 1.5 failures. a) probability that in a...
Failures in machine in an 8 hrs shift is 1.5 failures.
a) probability that in a period of 16 hrs, exactly 2 failures
take place.
b) probability that number of failures in 8 hrs exceed the
media (medium value) in more than 1.5 standard deviations
Homework Answers
Answer #1
Let X: failure in machines in an 8 hrs shift have exponential distribution with 1.5 failures.
a) P(X = 2)
For 16 hours the failure rate is 1.5 * 2 = 3 failures that is
The pdf of exponential distribution is,
The probability that exactly 2 failures takes place is 0.0074
b) P(X more than medium value)
The standard deviation of exponential distribution is,
Medium value is nothing but median.
The median of exponential distribution is,
The distribution function of exponential distribution is,
Probability that number of failures in 8 hours exceed the medium value is 0.422367
Not the answer you're looking for?
Similar Questions
What is the probability that a normal random variable will take a value that is less...
What is the
probability that a normal random variable will take a value that is
less than 1.05 standard deviations above its mean? In other words,
what is P(Z < 1.05)?
.8531
.1468
.9332
.0668
What is the
probability that a normal random variable will take a value that is
between 1.5 standard deviations below the mean and 2.5 standard
deviations above the mean? In other words, what is P(-1.5 < Z
< 2.5)?
.9938
.0668
.9270
.0730
What is...
3. A mix beverage machine releases a certain amount of syrup into a chamber where it...
3. A mix beverage machine releases a certain amount of syrup
into a chamber where it is mixed with carbonated water. The amount
of syrup follows a normal distribution with a mean of 1.29 fl.oz.
and a standard deviation 0.016 fl.oz.
a. Find the probability that a syrup amount does not exceed 1.33
fl.oz. (10)
b. Find the probability that a syrup amount is less than 1.29
fl.oz. (10)
c. Find the probability that a syrup amount exceeds the mean...
6. You have been provided the following probability distribution: Pets P(x) x*P(x) (x-µ)²P(x) 0 0.330 1...
6. You have been provided the following probability
distribution:
Pets
P(x)
x*P(x)
(x-µ)²P(x)
0
0.330
1
0.270
2
0.300
3
0.050
4
0.025
5
0.020
6 or more
0.005
Let x = the total number of pets in a
household
a. Complete table
b. What is the probability of a household having no
pets?
c. What is the probability of a household having exactly 3
pets?
d. What is the probability of a household having
between 1 and 3
pets?...
What is the probability that the following project will take more than 10 weeks to complete...
What is the probability that the following project will take
more than 10 weeks to complete if the activity means and standard
deviations are as shown below? (Round standard deviation to
2 decimal places and final answer to 4 decimal
places.)
Use Table B.
Activity
Mean
Standard
Deviation
1-2
5
1.3
2-3
4
1.0
1-3
8
1.6
The sales of medicines in a certain pharmacy behaves with a variable with a normal distribution...
The sales of medicines in a certain pharmacy behaves with a
variable with a normal distribution with means of 2000 pesos and
variance of 62500
a) What is the probability that the pharmacy will sell more than
1500 pesos?
b) How many standard deviations is a sales value of 3300 pesos in
the scope of the media? Can it be considered atypical (unusual) or
not? Why?
c) What is the 99th percentile of sales?
d) What is the probability that...
In an article in the Journal of Management, Joseph Martocchio studied and estimated the costs of...
In an article in the Journal of Management, Joseph Martocchio
studied and estimated the costs of employee absences. Based on a
sample of 176 blue-collar workers, Martocchio estimated that the
mean amount of paid time lost during a three-month period was 1.4
days per employee with a standard deviation of 1.4 days. Martocchio
also estimated that the mean amount of unpaid time lost during a
three-month period was 1.4 day per employee with a standard
deviation of 1.6 days. Suppose...
In an article in the Journal of Management, Joseph Martocchio studied and estimated the costs of...
In an article in the Journal of Management, Joseph Martocchio
studied and estimated the costs of employee absences. Based on a
sample of 176 blue-collar workers, Martocchio estimated that the
mean amount of paid time lost during a three-month period was 1.4
days per employee with a standard deviation of 1.3 days. Martocchio
also estimated that the mean amount of unpaid time lost during a
three-month period was 1.0 day per employee with a standard
deviation of 1.8 days. Suppose...
In an article in the Journal of Management, Joseph Martocchio studied and estimated the costs of...
In an article in the Journal of Management, Joseph
Martocchio studied and estimated the costs of employee absences.
Based on a sample of 176 blue-collar workers, Martocchio estimated
that the mean amount of paid time lost during a
three-month period was 1.3 days per employee with a standard
deviation of 1.9 days. Martocchio also estimated that the mean
amount of unpaid time lost during a three-month period was
1.4 day per employee with a standard deviation of 1.8 days.
Suppose...
Suppose that for a given computer salesperson, the probability distribution of x = the number of...
Suppose that for a given computer salesperson, the probability
distribution of x = the number of systems sold in 1 month
is given by the following table.
x
1
2
3
4
5
6
7
8
p(x)
0.04
0.10
0.13
0.30
0.30
0.11
0.01
0.01
(a) Find the mean value of x (the mean number of
systems sold).
Answer= 4.14
(b) Find the variance and standard deviation of x.
(Round your standard deviation to four decimal places.)
variance = 1.8604...
Question 1 The random variable X is a payout table of a casino slot machine. The...
Question 1
The random variable X is a payout table of a casino slot
machine. The probability mass function is given:
X -5 0 2
10 20 40 60
1000
Probability 0.65 0.159
0.10 0.05 0.02 0.01
0.01 0.001
Find the standard deviation of the payout .
Question 2
The random variable X is a payout table of a casino slot
machine. The probability mass function is given:
X -5 0 2
10 20 40 60
1000
Probability 0.65 0.159 ...
ADVERTISEMENT
Need Online Homework Help?
Get Answers For Free
Most questions answered within 1 hours.
ADVERTISEMENT
Active Questions
asked 5 minutes ago
asked 15 minutes ago
asked 24 minutes ago
asked 34 minutes ago
asked 48 minutes ago
asked 48 minutes ago
asked 48 minutes ago
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago