Question

7. Using the following, construct a complete frequency distribution include the mean, Variance, Standard Deviation. Also graph the Frequency Histogram, Polygon and Ogive. Indicate Modal Class, Median Class and describe the distribution (25 points) Frequency Class Limits 28 - 34 35 - 41 42 - 48 49 - 55 56 - 62 9. Find the weighted mean for three exams if the first one was worth 25% and the student received a 92, the second was worth 40% and the student received a 86 and the third was worth 35 % and the student received an 76? (3 pts) 10. It has been reported that 45 % of all cars on the highway are traveling at speeds in excess of 70 mph. If the speeds of eight random automobiles are measured via radar: (3pts cach 6 points) a. What is the probability that exactly 3 cars are speeding? b. What is the probability that at least one car is going over 70 mph 11. Using the probability distribution listed below, ( 3pts each 15 points) 0 0.10 1 .26 2 0.30 3 0.24 4 0.10 P(x) 0 a. Find the mean, variance and standard deviation b. Find P(x is no more than 3) c. Find P(x is at least 2) d. Find P(x<2 or x>2) e. Find P(x is less than 3) 12. It has been reported that 65 % of all internet users have been attacked by viruses. If 8 people are randomly surveyed, find the probability that exactly 5 have been attacked: (8 pts)

Answer 1

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6 Ginen, l 1 Probability that any car is travelling in excess of Tomph is given by Pop= 0.45 n=8 &q=1-P = 0.55 ② Pe ( exactly 03 cand are speeding) = since premains constant mith trials - there is only 02 possible outcomes re- speeding can or non-speeding con - # of trials are definite & independent disth with approximate binomial so we can n=8 & p=0.45 Now, P(x=3) where x urre- representing # of can speeding WILT, For binomial distr, Probe of & # of success in in trials is given by a - P(x-2) = ei pea 'Here, Succen • P(x-3) = &c, (0145€(0.537 being the speeding = 0.2568 Ann Pn (at least one can is over Tomph) = P( X21) - |- P(x=0) i. P(x=0) = 86 (0.45) Cooss) = 0.0084 • P(x21) - 1-0.0084 = 0.1916 An Wkt, for trials is o 1-2 ne. peq of can (6

11.)

@ Xo1|2|34| P(x) 0.10 0.26/0.30 0.24 010 {= since & P(Xi) = 1 : x is a probability dissbr. @ is mean = E(X) = {X. P(X) = 1.98 Am cim Var (X) = E(X) – (E(X))! E(X)= EXP(x) 5.22 . Var (x) = 5-22 - (1998) = 1:23 Am T(X)= vran(x) = 2123 = 1.14 Am © P ( x <3) = 1 - P(x=4) P(x=4) = 0.10 = 0.90 Am PCX53) = -0.10 - 0070 © Plat least 2) - P(x22) = P(x + 2) +P (X23)+8[*> +) & P(x=2) = 0.30 P(x=3) = 0.24 P(X= 4) = 0.10 {=00 0.64 Am & EXT) P(x+2) or P(X82) = P(x>2) + P(x(2) = 1-P(x=2) & P(x=2) = 0.30 in = 1-0.30 = 0.70 An

X	P(X)	X * P(X)	X^2	X^2 * P (X)
0	0.1	0	0	0
1	0.26	0.26	1	0.26
2	0.3	0.6	4	1.2
3	0.24	0.72	9	2.16
4	0.1	0.4	16	1.6
	1	1.98		5.22

(3) - P(X= 2) +P ( X = 1) +PLX=0) . P(X=2) 20.30 P(X=1) = 0.26 P(X=0) = 0.10 E=0.66 Am 13 Let & be the prob. that any Internet use have been attacked by virus be given by P o p=0.65 (given) n= Sample size =8 the can approximate binomial distro o trials are definite & Independent ire. No to users prob. of getting attacked by nerus is dependent 2 There are only on possible eether outcomes.lie being attacked or not. 7 p remains constant from trual to trial Pn( exactly s out of of have been attacked) = P(x=5)=? where in the su upusenting #of users being attacked WET, P(x=8) = "cape gre ; ein #of succes. | P(x=2) is prob. of . P(x=5) = 865 (065) (035) getting a success 10 in n trials. = 0.2786 Amm Here user being Sattacked by miru is treated as Success