Question #249368

1. Suppose that the algorithm, or robot reporter, typically writes proportion 0.65

of the stories on the site. If 15 new stories are scheduled to appear on a web site next

weekend, find the probability that

a) 11 will be written by the algorithm.

b) at least 10 will be written by the algorithm

c) between 8 and 11 inclusive will be written by the algorithm.

2. The amount of time it takes to assemble a computer is normally distributed, with a mean of 

50 minutes and a standard deviation of 10 minutes. 

a. What is the probability that a computer is assembled in a time between 45 and 60 minutes?

b. What is the probability that a computer is assembled in time more than 65 minutes?

c. What is the probability that their average time will be less than 45 minutes?


1
Expert's answer
2021-10-11T11:04:40-0400

Let x-number of stories written by robot. The distribution of the amount of the stories written by computer could be described as Binomial distribution with n = 15 and p = 0.65

1.

(a) P(x=11)=C(1115)(0.65)11(0.35)4=0.1792P(x=11) = C \binom{11} {15} * (0.65)^{11}*(0.35)^{4} = 0.1792

(b) P(x10)=P(x=10)+...+P(x=15)=0.2123+0.1792+0.1110+0.0476+P(x ⩾ 10 ) = P(x = 10) + ... + P(x=15) = 0.2123+0.1792+0.1110+0.0476 +

+0.0126+0.0016=0.5643+0.0126+0.0016 = 0.5643

(c) P(8x11)=P(x=8)+P(x=9)+P(x=10)+P(x=11)=0.1319+P(8⩽x⩽11) = P(x=8)+P(x=9)+P(x=10)+P(x=11)=0.1319+

+0.1906+0.2123+0.1792=0.714+0.1906+0.2123+0.1792 = 0.714

2.

N(50,10) = 50 + 10*N(0,1)

(a) P(45<50+10N(0,1)<60)=P(0.5<N(0,1)<1)=P(N(0,1)<1)P(45<50+10*N(0,1) < 60) = P(-0.5<N(0,1)<1)=P(N(0,1)<1)-

P(N(0.1)<0.5)=0.841340.30854=0.5328P(N(0.1)<-0.5)=0.84134-0.30854=0.5328

(b) P(50+10N(0,1)>65)=P(N(0,1)>1.5)=0.06681P(50+10*N(0,1)>65) = P(N(0,1)>1.5)=0.06681

(c) P(50+10N(0,1)<45)=P(N(0,1)<0.5)=0.30854P(50+10*N(0,1)<45) = P(N(0,1)<-0.5)=0.30854


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