Answer in Statistics and Probability for ABC #249908

From the following uniformly distributed data scene calculate the probability w.r.t statement.

The weekly output of a steel mill is a uniformly distributed random variable that lies between 110 and 175 metric tons.

1. Compute the probability that the steel mill will produce more than 150 metric tons next week.

2. Determine the probability that the steel mill will produce between 120 and 160 metric tons

Let “X” be a random variable representing the weekly output of the steel mill. First, we determine the “pdf” of “X” given as,

“f(x)=1/(b-a)” where “(a,b)” is the interval of the uniform distribution. For this case, this interval is, “(a=110,b=175)”.

Therefore,

“f(x)=1/(175-110)=1/65,space 110lt xlt 175”

“0,space elsewhere”

a.

The probability that the steel mill will produce more than 150 metric tons is given by,

“p(Xgt 150)=int^{175}_{150}(1/65)dx=(x/65)|^{175}_{150}=(175-150)/65=25/65=5/13”

Therefore the probability that the steel mill will produce more than 150 metric tons next week is “5/13” .

b.

The probability that the steel mill will produce between 120 and 160 metric tons is given by,

“p(120lt Xlt 160)=int^{160}_{120}(1/65)dx=(x/65)|^{160}_{120}=(160-120)/65=40/65=8/13.”

Thus, the probability that the steel mill will produce between 120 and 160 metric tons is “8/13.”