210 A random sample of 10 resistors is to be tested From pas

2.10 A random sample of 10 resistors is to be tested. From past experience, it is known that the probability of a given resistor being defective is 0.08. Let X be the number of defective resistors. a. What kind of distribution t unction would be recommended for modeling the r.v. X? b. According to the distribution function in (a), what is the probability that in the sample of 10 resistors, there are more than 1 defective resistors in the sample? 2.12 The consumption of maneuvering jet fuel in a satellite is known to be normally distributed with a mean of 10,000 h and a standard deviation of 1000 h. What is the probability of being able to maneuver the satellite for the duration of a 1-year mission? 2.13 Suppose a process produces electronic components, 20% of which are defective. Find the distribution of x, the number of defective components, in a sample size of five. Given that the sample contains at least three defective components, find the probability that four components are defective. 2.14 If the heights of 300 students are normally distributed, with a mean of 68 inches and a standard deviation of 3 inches, how many students have a. Heights of more than 70 inches? b. Heights between 67 and 68 inches? 2.16 Between the hours of 2 and 4 p.m., the average number of phone calls per minute coming into an office is two and one-half. Find the probability that during a particular minute, there will he more than five phone calls.

Solution

(2.10)(a) X follows Binomial distribution with n=10 and p=0.08

P(X=x)=10Cx*(0.08^x)*(0.92^(10-x)) for x=0,1,2,...,10

(b)P(X>1)= 1-P(X=0)

=1-10C0*(0.08^0)*(0.92^(10-0))

=0.5656115

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(2.12)1 years = 8760 hours

So the probability is

P(X>8760) = P((X-mean)/s >(8760-10000)/1000)

=P(Z>-1.24) = 0.8925 (from standard normal table)

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(2.13) X follows Binomial distribution with n=5 and p=0.2

P(X=x)=5Cx*(0.2^x)*(0.8^(5-x)) for x=0,1,2,3,4,5

So the probability is

P(X=4|X>=3) = P(X=4)/ P(X>=3)

=0.0064/0.05792

=0.1104972

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(2.14)(a) P(X>70) = P((X-mean)/s >(70-68)/3)

=P(Z>0.67) =0.2514 (from standard normal table)

So n=0.2514*300=75.42

(b) P(67<X<68) = P((67-68)/3 <Z< (68-68)/3)

=P(-0.33<Z<0) =0.1293(from standard normal table)

So n=0.1293*300=38.79

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(2.16)

Given X follows Poisson distribution with mean=2.5 per minute

P(X=x)=(2.5^x)*exp(-2.5)/x!

So the probability is

P(X>5) = 1-P(X=0)-P(X=1)-...-P(X=5)

=1-(2.5^0)*exp(-2.5)/1-...-(2.5^5)*exp(-2.5)/5!

=0.04202104