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Comprehensive Statistical Analysis and Probability Exam

Exam answers covering statistical analysis and probability concepts.

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    Comprehensive Statistical Analysis and Probability Exam A n s w er all 3 0 q u e s ti o ns. M a k e s ure your a n s w e r s are as c o m p l ete as poss i ble. Show all of your w ork and r e asoning. In p arti c ular, w hen there are calculati on s inv o l ved, you m ust s h o w how you c o me up w i th y our a n s w e r s w ith c r i tical w ork and/or nec e s s ary t a b l e s. A n s w e r s t h a t c ome s t raig h t f r o m progr a ms or soft w are pack a ges w i l l not be ac c epted. This e xam h a s 3 0 0 to t al p o i nts. Refer to the foll o wing table for Q u estio n s 1, 2 , and 3. S h ow all w o rk. Just the a nswer, wit h o u t su p p o rting work, will rec e i ve no credit. The t a ble s h ows tem p er a tures on the first 12 d a ys of October i n a s m all town in M a ryl a nd. Date Te m p eratu r e Date Te m p eratu r e Date Te m p eratu r e Oct 1 73 Oct 5 5 3 Oct 9 6 6 Oct 2 6 5 Oct 6 5 2 Oct 10 7 5 Oct 3 6 5 Oct 7 62 Oct 11 5 2 Oct 4 70 Oct 8 55 Oct 12 57 1 . Determine t h e fi v e - n u m b e r sum m ary f o r t h is d ata. (10 pts) 2 . Determine t h e m e an t e m p e rature. (3 p ts) 3 . Determine t h e m o d e(s), if an y . (2 p ts) 1. Obtain the five - number summary for the given data. The ordered series needed to obtain the five - number summary is given below. 52 52 53 55 57 62 65 65 66 70 73 75 Minimum = 52 First Quartile = Mean of the 3 rd and the 4 th terms = (53 + 55)/2 = 54.5 Median = Mean of the two middle terms = (62 + 65)/2 = 127/2 = 63.5 Third quartile = Mean of the 9 th and the 10 th terms = (66 + 70)/2 = 68 Maximum = 75 2. The mean temperature is,

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      12 1 1 1 1 73 65 57 745 62.0833 12 12 i i x X n = = =  + + + =  =  . 3. The mode(s) is (are) the highest frequent observation(s) in the given data. In the given data two observations 52 and 65 are repeated twice and the remaining are appeared only once. Therefore, the required modes are 52 and 65. Refer to the foll o wing fre q ue n cy distributi o n f or Questio n s 4, 5 , 6 , and 7 . S h ow all work. J u st t h e a n swe r , witho u t s u pporting w o rk, will re c eive no credit. The freq u ency d istribution below sh o ws the distr i bution for checkout time ( in minutes) i n UMUC Mi ni M a rt between 3 :00 and 4:00 PM o n a Frid a y afterno o n. Checko u t T im e ( i n minute s ) Frequ e ncy 1.0 - 1.9 6 2.0 - 2.9 7 3.0 - 3.9 2 4.0 - 4.9 3 5.0 - 5.9 2 4 . W h at perc e n tage o f the check o ut ti m es was l e ss t h an 5 m inutes? (5 p ts) 5 . Calculate t h e m e an o f this freque n cy distribution. (5 pts) 6 . Calculate t h e stan d ard de v iation of this freq u ency distribution. (10 pts) 7 . Ass u me t h at t h e s m all e st o b servation in this d ata s et is 1 . 2 m inutes. S u p p o se this o b servation w e re i n corre c tly recor d ed as 0 . 1 2 i nstead of 1 . 2. Will t he m e a n inc r e a s e , decrease, or r e m ain the sam e ? Will the median i n creas e , d ecrease o r r e m ain t he s a m e ? E x plain your answe r s. (5 p ts) 4. Obtain The percentage of the checkout times was less than 5 minutes, 6 7 2 3 18 90% 6 7 2 3 2 20 + + + = = + + + + 5. Obtain the mean of this frequency distribution: Checkout Time (in minutes) Frequency, f Midvalue, x f* x 1.0 - 1.9 6 1.45 8.7 2.0 - 2.9 7 2.45 17.15

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    3.0 - 3.9 2 3.45 6.9 4.0 - 4.9 3 4.45 13.35 5.0 - 5.9 2 5.45 10.9 Total 20 57 ( ) 57 The required mean 2.85 20 fx x f = = =   6. Obtain the standard deviation of this frequency distribution: Checkout Time (in minutes) Frequency, f Midvalue, x f*x f*x^2 1.0 - 1.9 6 1.45 8.7 12.615 2.0 - 2.9 7 2.45 17.15 42.0175 3.0 - 3.9 2 3.45 6.9 23.805 4.0 - 4.9 3 4.45 13.35 59.4075 5.0 - 5.9 2 5.45 10.9 59.405 Total 20 57 197.25 ( ) ( ) ( ) 2 2 2 Standard deviation 1 57 197.25 20 20 1 1.8316 1.3534 fx fx f s f − = − − = − = =     7. The mean decreases as the observation recorded incorrectly decreases the sum of the observations. The median remains the same as it is a measure of location not influenced much by the change in extreme observations. Refer to the foll o wing i nfor m ation f o r Q u estio n s 8 a n d 9 . S h ow all work. J u st t h e a n swe r , witho u t su p p o rting work, will rec e i ve no credit. A 6 - faced die is rolled t w o time s . Let A be the ev e nt th a t the outcome o f t h e first roll is greater than 4 . Let B be the event th a t the o u tcome o f se c o n d roll is an o dd n umber. 8 . W h at is t h e probabili t y t hat the o u tco m e o f t h e se c o n d roll is an o dd n u m be r , g iven that the first roll is great e r t h an 4? (10 pts)

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    ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) In the experiment of throwing a 6-faced die two times, the sample space is as follows, 1,1 , 1, 2 , 1,3 , 1, 4 , 1,5 , 1, 6 2,1 , 2, 2 , 2,3 , 2, 4 , 2,5 , 2, 6 3,1 , 3, 2 , 3,3 , 3, 4 , 3,5 , 3, 6 4,1 , 4, 2 , 4,3 , 4, 4 , 4,5 , 4, 6 5,1 , 5, 2 , 5,3 , 5, 4 , 5,5 , S = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 36 5, 6 6,1 , 6, 2 , 6,3 , 6, 4 , 6,5 , 6, 6 Let be the event that the outcome of the first roll is greater than 4. Let be the event that the outcome of second roll is an odd number. 5,1 n S A B A            =             = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )   ( ) , 5, 2 , 5,3 , 5, 4 , 5,5 , 5, 6 12 6,1 , 6, 2 , 6,3 , 6, 4 , 6,5 , 6, 6 1,1 , 1,3 , 1,5 , 2,1 , 2,3 , 2,5 , 3,1 , 3,3 , 3,5 , 4,1 , 4,3 , 4,5 , 18 5,1 , 5,3 , 5,5 , 6,1 , 6,3 , 6,5 5,1 , 5,3 , 5,5 , 6,1 , 6,3 , 6,5 6 The probability th n A B n B A B n A B      =           =  =        =   = ( ) ( ) ( ) at the outcome of the second roll is an odd number, given that the first roll is greater than 4 is, 6 1 | 12 2 n A B P B A n A  = = = 9 . Are A and B inde p en d en t ? W h y o r why n o t? (5 p ts) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 From the Question 8 , we have | 2 Obtian the probability of the event that the outcome of second roll is an 18 1 odd number, i.e. . 36 2 1 Since | , the events and are independent. 2 P B A n B P B n S P B A P B A B = = = = = = Refer to the fol l owi n g data to answer q uesti o ns 1 0 a n d 11 . Show all work. Just the a nswer, wit h o u t s u p p orting w o rk, will re c e i ve n o cr e dit. A rand o m sample of ST A T2 0 0 we e kly study times in h o urs is a s foll o ws: 2 14 1 5 1 8 20 10 . Fi n d the sta n dard d eviation. (10 pts)

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