Question

- Question 8 a) A green die is biased so that a 6 is 5 times as likely to appear as any other number, and that the five other outcomes are equally likely. Leaving your answer in exact form, what is the probability of a 6 appearing in a single toss of the green die? 5 points 0.5 b) A red die is biased so that the probability that a 1 appears in a single toss is Let X be a random variable that represents the number of times 1 appears in 170 tosses of the red die. Leaving your answers in exact form, write the values for E(X) and Var(X) E(X) = and Var(X) = c) Let Y~B(400, 3/10). Using the normal approximation with continuity correction, we can write P(170 <Y <190) > Pla<W<b) where W-N (E(X), Var(X)) for a= D and b= 0 d) Let C N (30,9). Then P(29<C<32) = P(m<Z<n) where 2 is the standard nomal variable and the exact values for m and n are m = O and n=

Answer 1

SWASTIK Date : Page No. sophea) It ia given that the die in biased and 6 iq 5 times appear as quy other number ie if we throw this die n times then lora or 3 or 4 or 5 will Lappeter x (gay) times then 6 appear 5x times. then consider the total number of tooges n can be split as - No. of six + No of (one or two or three or four or five) =n : 5% + ( x +x+x+x+2) = n 5% 75% on 103 = n a = 10 so No. of 6 will be in in times = 5 xhan and no of 1 or 2 or 3 or 4 or 5 in intimy na we know that no. of 6 in ntime - 1/2 and probability of getting 6 in intimes = 12 which will be equal to probability of getting . 6 In One to 88 because it is Independent of n Probability of a 6 appearing in a top & = 2 2 = 0,5 so

Date : Page No. It is given that probability of getting 2 18 3 re p = 2/3 ( probability of getting sucess) If this process of experiment will be done repeatedly 170 times (toases then it will follow a blnomial u distribution and a be the number of times I appear go the pim. flo Pex=x) = 170 call x=0,1,2,- ---170. since we know that if an Ben, p) co E (Xenpe VCX) 2 M HClop and Here KNB(170,22 80. - E (A)2 170x2 = 510 8 = 63.75 V(x) = 170 x 35 u Q = 39.84 Criven that YN BC 400, 86). then P1170242190). = 9 we know that if x is a dycrete random variable and we have to find that

SWASTIK Date : Page No. s Plac XL b) then by continuity correction Plack<b) n planda <x< b + 12) go P1170€ 46190) P1170- <Y€ 190 + $ 2 = P(169.5 LYS 190.5) -0 Now, because 'YN BC 400, 3) then E(Y= 400x = 120. U(Y) = 400 x 3 x 7 = 84. 80 from PL 1702 45 190) == Pl 169.5-120 < Y-120 < 190.5-120) - - P45,40 %'WE 769) So Q = 5,40 iba 7,63.

SWASTIK Date : Page No. since CNN1.30,9). So . . PL 29 C232) = P1 19-302 C-30 (32-30) E = 5.2<5). on comparing we get m = -1, n = 4.