Question

Question 02: Past data from a soccer league suggests that 6% of new players sustain a serious injury within the first year of their arrival in a new club. Leeds United recruited 13 new payers this season. a. What is the probability that none of these players will sustain any serious injury this year? (1 Mark] b. What is the probability that exactly two of these players will sustain a serious injury this year? (1 Mark] C. Compute the mean and standard deviation of this probability distribution. [1 Mark] Question 3: The Winter snowfall in Toronto follows a uniform distribution, and ranges between 32 cm and 216 cm. a. What is the average snowfall in Toronto during the winter? What is the standard deviation? [2 marks] b. What is the probability of less than 120 cm of snow during winter? [1 Mark] Question 4: The average US adult spends 38 minutes per day on Facebook. If the time they spend on Facebook follows a normal distribution and the standard deviation is 5.6 minutes, what is the probability that a particular user will spend - a. More than 60 minutes per day on Facebook. [1 Mark] b. Less than 10 minutes per day on Facebook. (1 Mark] C. Between 20 and 30 minutes per day on Facebook. (1 Mark) Question 5: According to research, the typical TikTok user spends 316 hours per year using the app. Hitesh Makhija is a big TikTok fan, and he uses the app whenever he has some free time. Assume the number of hours spent in TikTok follows a normal probability distribution with a standard deviation of 11.5 hours. If Hitesh is in the top 2% in terms of time spent on TikTok, how many hours did he spend in TikTok last year? [3 Marks]

Answer 1

Quesa vustaining serious injury Let X denote no. such that such that P ( X = n ) of players Xwl Binomial (113,0.06.) Consj 10.06).*.40:14) 13-1 4 x 20.12.13 P ( X = None ] = P(x=0) = (3) 10.06)°/0°94)13 = 0.4474 16 © PL Exacity 2) - P(X=2) = ($ 260.0634 60.94)"} 0:1422 - Mean €(x) = np = 13x0.06_= 0.78 Standard Deviation ='secx) = npcl-p) 13 x0.06x0.94 10.8563 Ques 3. in Toronto such that het y denote winter Snowfall Yn Uniform ( 32'9216) So, fyly) 32242216 216-32 184 a o 124 cm 2 = 12 ECY) = Average bta +216 +32 = Standard Deviation SD(Y) = |(6-2)2 = (216-32) 1 2 53•1168 b. for Uniform, Fyly ) = P(YLY) = 7-a So, P/X220) - Fy (120) = 0.4783 Б. 120-32 216-32

= P(Z >3093) 5.6 5.6 1 P =_316 + 11:5 x2:05421339:618 Thoms Que 4. Let X v Normal (ut = 38 , r = 5.6) © P[ X760] P[x - 38 > 60 - 38 U p[x<10] - P[ *=38 } <10=38 1 = • P/ 26.5 © P[ 20< x < 30 ] - P[ 20-38 < 7 2 30-38 P[-321422 Z 6-1:4286] =P ( 33–104286) – PLZ <=312142 0.0766 0.0754 Here, N(o,) Let yn Normal (3169 5] We need to find y such that P{y ry] = 0.02 P{ycy = 0.98 y-316 = 0.98 1165 Plz ez 2* )= 0.98 We need to find z-score z z* =y 9-316 such that area under curve is Using normal Table , + 3 * = 9-316 0.0007 z = X-38, 5.6 1105 Ques 5. PL g -316 11.5 it 1105 0.98 = 2:054 llos of