randomly selected Disney World visitor is less than one standard deviation above

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]

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