Make sure it\'s 100% correct plz 6 [2.5 points] A manufacturer knows that, on av

Question

Make sure it's 100% correct plz 6 [2.5 points] A manufacturer knows that, on average, 2% of items manufactured will have de- 0-39 fects. Use the normal approximation to the binomial distribution to determine the probability 0.8770 that among 200 items, (a) at the most 5 will be defective; 09279 0.9616 (b) anywhere from 4 to 7 inclusive will be defective. 0.9693 9736 0.9 41 0.9932 0 0.9011 0.9913 /-97 0.9972 a 0.9994 0.9992 0.9999 O.9989 0.9985 0.9986 0.9989 0.9980 7 0.9997 0.9997 70.9997 0. 0.9995 0.9995 0 0.9990 0.9993 0.9997 0.9997 0.9998 //(ooooo oooo was o 997 995 996 L S4 87 13 34 51 63 73 80 86 90 93 95 O. O. O. O. O. O. O. O. O. O. 0. 00 00.0 0 6 67002 9 ssi 2 9229 54 89 04 95 3 467 79 85 89 00000 00000 O. O. O. O. O. O. O. 00 O. O. O. 0. O. O. 0.

Explanation / Answer

Using binomial formulae,

E(x) = np = 0.02 * 200 = 4

Var(x) = np(1-p) = 0.02 * 0.98 *200 = 3.92

a)

p(x<=5) = p(x<5.5) ( using continuty correction , p(x<=n) = p(x< n+0.5))

= p( z < 5.5 - 4 / sqrt(3.92) )

= p( z < 0.7576)

= 0.77565

b)

p(4<=x<=7) = p(3.5<x<7.5)

= p(x<7.5) - p(x<3.5)

= p(z<7.5 -4/sqrt(3.92)) - p(z< 3.5 - 4 / sqrt(3.92))

= p(z< 1.7678) - p( z< -0.25254)

= 0.96145 - (1-0.59969)

= 0.56114

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