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Suppose that lifetimes of light bulbs produced by a certain company are normal random
variables with mean 1000 hours and standard deviation 100 hours. Suppose that
lifetimes of light bulbs produced by a second company are normal random variables
with mean 900 hours and standard deviation 150 hours. A person buys one light bulb
manufactured by the first company and one by the second company. What is the
probability that at least one of them lasts 980 or more hours?
The simplest error detection scheme used in data communication is parity checking.
Usually messages sent consist of characters, each character consisting of a number of
bits (a bit is the smallest unit of information and is either 1 or 0). In parity checking,
a 1 or 0 is appended to the end of each character at the transmitter to make the total
number of 1's even. The receiver checks the number of 1s in every character received,
and if the result is odd it signals an error. Suppose that each bit is received correctly
with probability 0.999, independently of other bits. What is the probability that a
7-bit character is received in error, but the error is not detected by the parity check?
(Hint: use the Binomial distribution)
Negation of statement (A ∧ B) → (B ∧ C) is: (A ∧ B) →(~B ∧ ~C)
Can you explain the solution please?
The production department of WSS electronics wants to explore the relationships between the number of employees who assemble a certain product and the number of units produced per hour. The complete set of paired observations follows:
No. of employees: 2,4,6,8,5,12,10,9
Productions (units): 20,15,25,28,22,30,35,40
Q: Given Email B with its feature vector. Compute the probability of email B being “spam” and “ham” using Naïve bayes algorithm and then finally assign class label (spam or ham).
P(spam) = 0.65
Email B = < 0, 1, 1, 1 >, = < count(meeting), count(enron), count(dating), count(hi) >
P (meeting | spam) = 0.6, P (meeting | ham) = 0.02
P (enron | spam) = 0.4, P (enron | ham) = 0.001
P (dating | spam) = 0.7, P (dating | ham) = 0.005
P (hi | spam) = 0.3, P (hi | ham) = 0.09
1. The owner of a theme park would like to know the average amount of time visitors spend there. To do this, he asked 25 of them and found out that they stayed for an average of 123.5 minutes with a standard deviation of 10.5 minutes. With a confidence level of 95 percent, determine the confidence interval estimate of the population mean.
Suppose we wish to test Ho: µ = 47 versus Ha: µ > 47. What will result if we conclude that the mean is greater than 47 when its true value is really 52?
Q: Suppose a dataset has 8500 email collection. Among 8500 emails, 4000 emails are not-spam and remaining are spam emails. The word “dating” is used as a feature, whose frequency/count in spam emails are 310 and 106 in not-spam emails. You have to compute two probabilities using bayes theorem, only knowing it contains the word “dating”.
First: Probability of an email being spam? Second: Probability of an email being not spam?
Q: A certain virus infects one in every 20 people. A test used to detect the virus in a person delivers positive outcome at 85% accuracy for infected persons. Moreover, it provides negative outcome for healthy persons at 95% accuracy. Compute followings:
a) Find the probability that a person has the virus given that his test outcome is positive.
b) Find the probability that a person does not have the virus given that his test outcome is negative.
c) Find the probability that a person has the virus given that his test outcome is negative.
3. A diet clinic states that there is an average loss of 24 pounds for those who stay on the program for 20 weeks. The standard deviation is 5 pounds. The clinic tries a new diet, reducing salt intake to see whether that strategy will produce a greater weight loss. A group of 40 volunteers loses as as average of 16.3 pounds each over 20 weeks. Should the clinic change the new diet? Use a 0.05
