Question
randomly selected Disney World visitor is less than one standard deviation above the mean. 5.58 Public Policy and Political Science Approximately 100 children's products are recalled every year^12. In particular children's clothing is recalled for a variety of reasons, for example, drawstrings that are too long and pose a hazard, small buttons that may break off and cause choking, and material that fails to meet federal flammability standards. suppose the nu,ber of recalls of children's clothing during a gievn month is a random variable with probability distribution given in the table below. a. Find the mean, variance, and standard deviation of the number of recalls of childre,s clothing during a given month. Suppose the nu,bner of recallsw in a given month is at least three. What is the probability that the number of recalls that month will be at least five? C. if the number of recalls in a given month is above more than one standard deviation from the mean, the federal gievernment issues a special from the mean, the federal government issues a special warnin directed toward parents. What is the pobability that a special warning will be issued during a given month?
Explanation / Answer
X denotes the number of recalls in a month
the distribution of X is given as
x: 0 1 2 3 4 5 6
p(x): 0.005 0.185 0.275 0.305 0.200 0.020 0.010
a) hence mean E[X]=0*0.005+1*0.185+2*0.275+3*0.305+4*0.2+5*0.02+6*0.01=2.61
variance=V[X]=E[X2]-E2[X]
now E[X2]=0*0*0.005+1*1*0.185+2*2*0.275+3*3*0.305+4*4*0.2+5*5*0.02+6*6*0.01=8.09
hence variance=8.09-2.612=1.2779 [answer]
standard deviation is s=sqrt(1.2779)=1.130 [answer]
b) the number of recalls is=n a given month is at least 3.then the probability that the number of recall would be at least 5 is
P[X>=5|X>=3]=P[X>=5 and X>=3]/P[X>=3]=P[X>=5]/P[X>=3]={P[X=5]+P[X=6]}/{P[X=3]+P[X=4]+P[X=5]+P[X=6]}
={0.02+0.01}/{0.305+0.2+0.02+0.01}=0.0561 [answer]
c) one standard deviation from mean means E[X]+s*1=2.61+1.130=3.74
the federal govt issues a special warning directed towards parent if X>3.74
so the probability of that is
P[X>3.74]=P[X=4]+P[X=5]+P[X=6]=0.2+0.02+0.01=0.23 [answer]