Julia Deutsch _ HW_ Week 3 (1)

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Jan 9, 2024

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Julia Deutsch HW: Week 3 STATISTICS AND DATA ANALYSIS HOMEWORK EXERCISES PETER LAKNER 3. A normal random variable X has mean 3.0 and standard deviation 0.2. What is the probability that X falls between 2.75 and 3.1? IN R: > a=pnorm(3.1,3,0.2) > b=pnorm(2.75,3,0.2) > a-b [1] 0.5858127 The probability that X falls between 2.75 and 3.1 = 0.5858127 ≈ 59% 4. Suppose that X follows normal distribution with mean 5.5 and standard deviation 0.3. Find a number w such that X < w with 30% probability. IN R: > qnorm(.3,5.5,0.3) [1] 5.34268 P(X < 5.34268 ) = 30% 9. You are considering a quality inspection scheme to use on the spark plugs which are sent from your supplier. These spark plugs come in a shipments of 50,000. Denote the unknown proportion of defective spark plugs in the shipment by p. Ideally you would like to reject the shipment if p > .05 and accept it if p ≤ .05. In practice you can’t follow this plan since you don’t know p. Instead you decide to apply a scheme that consists of the following steps: A random sample of 20 of the spark plugs will be selected from each shipment. Each of the selected plugs will be tested to see whether it is defective or not. (The test involves measuring the plug gap and determining the electrical resistance.) You will note as X the (random) number of defective plugs in the sample. If X < 2 then the shipment passes your quality standard. If X ≥ 2 then the shipment fails the quality test and will be returned to the supplier.

(a) Find the probability that the shipment is rejected when p = .05 (this corresponds to an “error” since at p = .05 we would want to accept the shipment). P(X ≥ 2) = 1-[P(X=0)+P(X=1)] IN R > a=dbinom(0,20,0.05) > b=dbinom(1,20,0.05) > 1-(a+b) [1] 0.2641605 P(X ≥ 2) = 0.2641605 ≈ 26% (b) Find the probability that the shipment is accepted when p = .1 (this corresponds to an “error” again since at p = .1 we would want to reject the shipment). p(x<2)=p(x=0) + p(x=1) p(x<2)=dbinom(0, 20, .1)+dbinom(1, 20, .1) IN R: > c=dbinom(0,20,.1) > d=dbinom(1,20,.1) > c+d [1] 0.391747 P(X<2) when p= .1 0.391747 ≈ 39% (c) Find the probability that the shipment is accepted when p = .2. p(x<2)=p(x=0) + p(x=1) p(x<2)=dbinom(0, 20, .2)+dbinom(1, 20, .2) IN R: > e=dbinom(0,20,.2) > f=dbinom(1,20,.2) > e+f [1] 0.06917529 P(X<2) when p= .2 0.06917529 ≈ 7%

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