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The number of hits daily is a normally distributed random variable with a mean of 10,000 and a standard deviation of 2,400. a. What is the probability of getting more than 12,000 hits?


a school administrator claims that less than 50% of the students of sinapangan national high school are dissatisfied by the community canteen services.the claim by using sample data obtained from a survey of 500 students of the school where 54% indicated their dissatisfaction of the community canteen service.use a=0.05

Consider a random process X(t) de ned by

X(t) = U cos t + (V + 1) sin t; 􀀀-1 < t < 1

where U and V are independent r.v.'s for which

E[U] = E[V ] = 0; E[U2] = E[V 2] = 1

(a) Find the autocovariance function CX(t1; t2) of X(t).

(b) Is X(t) WSS?


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)


activity 2 solve the following problems. show all your works. 1 in a study of the life expectancy of 500 people in a certain geographic region, the mean age at death was 720 years, and the standard deviation was 5.3 years. if a sample of 50 people from this region is selected, find the probability that the mean life expectancy will be less than 70 years.

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?


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)


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


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