MTH 302 long questions solved by Pisces girl My Lord! Increase me in knowledge.

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1 MTH 302 long questions solved by Pisces girl My Lord! Increase me in knowledge. Question No: 5 ( Marks: 3 ) In a school, 50% students study science subjects and 30% of them study biology. What is the probability that the student studies Biology? Soluion: 30/100 3/10 30% 0.3 Question No: 7 ( Marks: 5 ) A student goes to the library. The probability that she checks out (a) a work of fiction is 0.40, (b) a work of non-fiction is 0.30,, and (c) both fiction and non-fiction is What is the probability that the student checks out a work of fiction, non-fiction, or both? Solution: Probability that she checks out for work of fiction = 2/5 Probability that she checks out for non-fiction = 3/10 Probability that she checks out for both = 1/5 Probability that the student checks out for fiction, non-fiction, or both = 2/5 + 3/10 + 1/5 = 9/10 Question No: 8 ( Marks: 5 ) Calculate the mean, median and mode for the following set of data 2,2,1,4,4,8,5, 6, 8, 19, 2, 1, 6, 25 Mean : Efx/n=93/14= 6.64 (14)

2 Median: 5+6/2=11/2=5.5 Mode : 2 Question No: 10 ( Marks: 10 ) Find all the quartiles for the following data set: 4.3, 5.1, 3.9, 4.5, 4.4, 4.9, 5.0, 4.7, 4.1, 4.6, 4.4, 4.3, 4.8, 4.4, 4.2, 4.5, 4.4 Solve: Arrange this data in Acessending order: Formula of 1st Q: N+1/4 17+1/4 18/4 4.5 Since we get position of first quartile as decimal fraction, so we proceed as follows Q1=2nd values+0.5*(3rd value-2nd value) *( ) (0.1) 4.15 Position 2nd Q 2*18/4=9 Q2=5th value 4.3 Poistion of 3rd Q 3*18/4=13.5 Q3=7th value+0.5*(8th value-7th value) ( ) 4.9

3 Question No: 14 ( Marks: 3 ) Define median and find the median of the following set of data 3, 6, 11, 14, 19, 19, 21, 24, 31 Solve Gives Median or Number in Middle Median is =19 Question No: 16 ( Marks: 3 ) Write the formula of seasonal variation. Solve Calculate Actual trend for each period Question No: 23 ( Marks: 2 ) In which condition none of the hypothesis testing procedures can be safely used. Solve If underlying population is not normal and we have a small sample Then none of the hypothesis testing procedures can be safely used. Question No: 24 ( Marks: 3 ) If the difference between two sample means is 11.29, n1 = 30 and n2= 50, then find the Standard Error of Difference in Sample Means (STEDM) Solve Standard Error of Difference in Sample Means (STEDM) = 11.29(1/30 + 1/50)^1/2 = 2.60 (B.P331) Question No: 33 ( Marks: 3 )

4 What are the differences between Normal distribution, Binomial Distribution and Poisson distribution? Solve A standard normal distribution is a distribution with mean = 0 and standard deviation = 1. The Y-axis gives the probability values. The X-axis gives the z (measurement) values. Each point on the curve corresponds to the probability p that a measurement will yield a particular z value (value on the x-axis.). Returns the individual term binomial distribution probability. Use BINOMDIST in problems with a fixed number of tests or trials, when the outcomes of any trial are only success or failure, when trials are independent, and when the probability of success is constant throughout the experiment. The Poisson distribution is most commonly used to model the number of random occurrences of some phenomenon in a specified unit of space or time. For example, The number of phone calls received by a telephone operator in a 10-minute period. The number of flaws in a bolt of fabric. The number of typos per page made by a secretary. Question No: 34 ( Marks: 3 ) Define normal distribution Solve A standard normal distribution is a distribution with mean = 0 and standard deviation = 1. The Y-axis gives the probability values. The X-axis gives the z (measurement) values.

5 Each point on the curve corresponds to the probability p that a measurement will yield a particular z value (value on the x-axis.). Question No: 57 ( Marks: 3 ) Find the single discount rate that is equivalent to the series18%, 12%, 9%. Solution: Let list price = 100 Series trade discount: Netprice = List price(1-d1)(1-d2)..(1-dn) = 100 (1 -.18)(1 -.12)(1 -.09) = 100(.82)(.88)(.91) =100(.657) =65.7 Single discount rate = = 34.3% Question No: 58 ( Marks: 5 ) Compute the amount of compound interest for Rs 3500 at 6% per annum for 2 years. Solution: S = P(1+R/100)^n S= Money accrued after n years also called compound amount P= Principle amount R= Rate of interest n= number of periods

6 S= 3500 (1 + 6/100)^2 =3500 (1 +.06)^2 =3500(1.06)^2 = 3500(1.1236) S=3933 Compound interest = S P = = 433 Question No: 59 ( Marks: 10 ) It is advised by the exam department of an institute to the assistant to develop a structure of marks obtained by three students in three different subjects for very quick view & analysis with the following instructions to be followed strictly as; a) marks obtained by EACH student in EVERY subject b) marks in EACH subject be obtained by EVERY student. Also describe & justify the optimality of the structure. If such model is possible then suggest its further extension for numerous numbers of students. Solution: The optimal method to solve this problem is by using matrices Let the student names be: 1. Ahmed 2. Shahzad 3. Amjad Let the subjects be: 1. Math 2. Chemistry 3. Physics

7 Math Chemistry Physics Ahmed Shahzad Amjad Now let s convert the above table into a matrix: A = BENEFITS OF MATRICES: The below benefits can help in further extension 1. Can handle large amount of data: So w.r.t to the above question you can increase the no. of students & subjects as per your requirement and a 2. Analysis of data: By using different matrix functions you can easily find out the respective averages and the grades of students. 3. Simple & Easy: The above mentioned matix structure is simple and easy to understand and the data can be easily converted from table to matrix 4. Easy to evaluate performance: Its easy to find each students performance in a specific subject 5. Easy extension: As matrices can handle large amounts of data so you just need to enter the data rest is done by matrices functions.

8 Question No: 63 Write which of the given function is used for which of the below given tasks. Functions: RATE(total number of payment periods, payment made each period,pv,fv,type,guess) PPMT(interest rate, period, total number of payment periods, pv, fv, type) Question No: 64 ( Marks: 2 ) Define frequency with example. Answer: The number of times a certain value or class of values occur. Question No: 65 ( Marks: 2 ) Define Type-I error Answer: Statistical probability in hypothesis testing that the test sample supports a conclusion that a value is misstated when, in fact, the value is correctly stated. Defined also as an incorrect decision to reject something that should have been accepted, it is the mirror image of type 2 error. If the level of significance (called alpha level) for a test result is set too high, the possibility of type 1 error is reduced but the possibility of type 2 error is raised by the same factor, and vice versa. Also called alpha error or alpha risk. Question No: 66 ( Marks: 2 ) A coin can be tossed in 3 ways. A die can be thrown in 6 ways. A coin and a die together can be thrown in 18.ways. Question No: 69 ( Marks: 3 )

9 Explain negative binomial distribution. Answer: It returns the probability that there will be number_f failures before the number_s-th success, when the constant probability of a success is probability_s. This function is similar to the binomial distribution, except that the number of successes is fixed, and the number of trials is variable. Like the binomial, trials are assumed independent. Question No: 72 ( Marks: 5 ) a) Write all the combinations of abcd taken 1 at a time. Answer : 4!/1!(4-1)! = 4 b) Write their combinations taken 2 at a time. Answer : 4!/2!(4-2)! = 4.3.2/2.2 = 6 c) Write their combinations taken 3 at a time. Answer : 4!/3!(4-3)! = 4.3.2/3.2 = 6 d) Write their combinations taken 4 at a time. Answer : 4!/4!(4-4)! = 1 Question No: 73 ( Marks: 10 ) Find the standard deviation and variance for 10 randomly selected data values : 44, 50, 38, 96, 42, 47, 40, 39, 46, 50. Answer: Standard Deviation = Question No: 74 ( Marks: 10 ) Find the mean, median, mode, and range for the following list of values: 13, 18, 13, 14, 13, 16, 14, 21, 13

10 Answer: Mean = 15 (N=9) Median = 14 Mode = 13 Range = 13 Question No: 75 ( Marks: 2 ) Define the Null Hypothesis. ANS: Null hypothesis is a scenario which explain a given set of data. It is tested to determine whether data provides sufficient reasons to pursue some alternative hypothesis. It is a hypothesis that states there is no difference between 2 or more sets of data. Question No: 76 ( Marks: 2 ) A coin can be tossed in 3 ways. A die can be thrown in 6 ways. A coin and a die together can be thrown in 3 X 6 = 18.ways. Question No: 77 ( Marks: 2 ) Find harmonic mean (HM) of 10,12,14,17. ANS: Harmonic mean,hm = n/(1/x1 + 1/x2 + 1/x3 + 1/x4 + +1/xn) N= 4 X1 = 10 X2 = 12 X3 = 14 X4 = 17 1/x1 + 1/x2 + 1/x3 + 1/x4 = 1/10 + 1/12+ 1/14+ 1/17 = Harmonic mean = 4/0.2385= Question No: 79 ( Marks: 3 )

11 Given for a frequency distribution mode = 18, mean = 21.Calculate median. Using these values comment on skewness of distribution. ANS: We know that, Since, So, Mean-mode= 3(mean- median) Median= (2*mean+mode)/3 Mean= 21 Mode= 18 Median = (2*21+18)/3 = (42+18)/3 = 60/3 = 10 The distribution is Moderately skewed and unimodal distribution. Question No: 80 ( Marks: 3 ) How many different ways can you select 2 letters from the set of letters: X, Y, and Z? (Hint: In this problem, order is NOT important; i.e., XY is considered the same selection as YX.)ANS: Total number of letters = 3 Letters taken at a time = 2 Number of ways in which 2 letters can be selected out of 3 = 3C2 = 3! / 2!(3-2)! Question No: 81 ( Marks: 5 ) =3*2*1/ 2 How many possible permutations can be formed from the word MATHEMATICS? ANS: Total number of alphabets in mathematics = 11 M = 2 A = 2 T = 2 = 3

12 H = 1 E = 1 I = 1 C = 1 S = 1 Permutations = 11! / 2! * 2! * 2! *1! * 1! * 1* 1! * 1! = 11! / 8 = 4,989,600 Question No: 82 ( Marks: 5 ) Find the standard deviation of 4, 9, 11, 12, 17, 5, 8, 12, 14 ANS: Standard deviation for sample = sqrt[sum(x- )2/n-1) = mean = ( )/9 = 92/9 = n= 9 so, n-1 =9-1 = 8 X X-X (X-X)

13 2 = Standard deviation = = 4.18 Question No: 83 ( Marks: 5 ) In a normal distribution what proportion of cases will fall between 20 and 35? Question No: 84 ( Marks: 10 ) Form a regression line from the data below. Departments X Y l ANS: Slope,b = [n*sum(x*y) sumx*sumy]/ [n*sumx2- (sumx)2] intercept, a = (sumy b*sumx)/n n= 7 Departments X Y X2 X*Y

14 sumx =272 sumy= 413 sumx2 = sumx*y = slope, b = [7(13509)-272*413]/[7(15138)-(272)2] = -17,773/31,982 = intercept, a = [413-(-0.55)272]/ 7 = equation of regression line is given by, Y = a+ bx Y = x Question No: 85 ( Marks: 10 ) The following data gives the height (in inches) of eleven 9-years old boys in a primary school. 57, 52, 51, 49, 55, 54, 50, 48, 53, 56, 47 a) Find first, second and third quartiles. b) Find interquartile range, Quartile deviation. ANS:a) Data in arranged order: 47,48,49,50,51,52,53,54,55,56,57 number of data points, n = 11 position of Qi = i(n+1)/4 Position of Q1= (n+1)/4 = (11+1)/4 = 12/4= 3 So, Q1= 3rd value = 49

15 Position of Q2= 2(11+1)/4 = 24/4 = 6 So, Q2= 6th value = 52 Position of Q3= 3(11+1)/4 = 36/4 =9 So, Q3= 9th value 55 b) Interquartile range = Q3-Q1 = = 6 Quartile deviation = (Q3-Q1)/2 = (55-49)/2 = 6/2 = 3 Question No: 135 ( Marks: 2 ) What will be y-intercept b in the regression line Y=aX+b if Y = 9, X = 3 and a = 2? Solution: 9 = (2 x 3) + b b = 9 6 b = 3 will be the Intercept Question No: 136 ( Marks: 3 ) What are the differences between Normal distribution, Binomial Distribution and Poisson distribution? Question No: 137 ( Marks: 3 ) Write the formula of seasonal variation.

16 Solution : Seasonal variation is a component of a time series which is defined as the repetitive and predictable movement around the trend line in one year or less. It is detected by measuring the quantity of interest for small time intervals, such as days, weeks, months or quarters. Hence Season variation = Actual Trend Question No: 140 ( Marks: 5 ) Calculate the S.D of the following data 2, 4, 4, 5, 10, 16, 17 Solution: X is the mean which is 8.28 X X-X (X-X) N = 7

17 X = 58/7= /2 S.D = SUM (x-x)2 ========= n ====== 6 = 6.12 is the answer Question No: 151 ( Marks: 2 ) What will be the correlation coefficient r between variables X and Y if varx=4,vary=9 and Cov(X,Y)=3? Answer R=9 Question No: 152 ( Marks: 2 ) What are the disadvantages for the so larger & so smaller values of smoothing constants in forecast analysis?

18 ANSWER : A small value will have less of a smoothing effect and be more responsive notice that this technique same disadvantages as the simple moving average as larger values actually reduced the level of smoothing effect. Question No: 153 ( Marks: 2 ) What will be the standard deviation of a sample of size 25 if its population variance is 2. ANSWER Standard deviation = 12.5 Question No: 154 ( Marks: 3 ) A random sample of size n is drawn from normal population with mean 6 and standard deviation 1.2; if z = 4, 8 what is n? ANSWER : N=2 Question No: 155 ( Marks: 3 ) Find Covariance Cov(X,Y) if var(x)=7,var(y)=6 and correlation coefficient r between X and Y is 0.3? ANSWER COV(X,Y)=1.3 Question No: 158 ( Marks: 5 )

19 Using the formula find the coefficient of linear correlation with the help of given table X Y ASNWER : X Y x2 y2 xy Put the values in the formula R= 4(234)-(28)(32) (4(206)-(784))*1/2*(4(266)-(1024))*1/2 = ( )*1/2*( )*1/2 = 976 (40)*1/2*(40)*1/2 = 40 20*20 r= 0.10

20 Question No: 159 ( Marks: 5 ) According to a survey, a certain city has a population of 100,000 people age 22 and over with standard deviation of 49. Of them 60% are married. If we draw a sample of 1600 people then find the chance that 58% or less of the people are married.