An analog signal received at a detector (ASD) has strength which has a normal di
ID: 3323049 • Letter: A
Question
An analog signal received at a detector (ASD) has strength which has a normal distribution with = 95 microvolts and a standard deviation of =5.7 microvolts. Suppose that we measure 456 independent analog signals at the detector. Let x be the mean strength and X be the random variable representing the mean strength and let Xtot be the random variable representing the sum of the strengths of the 456 received signals.
a) About what proportion of ASD's have a strength between 90 and 100 microvolts?
b) About proportion of ASD's have strengths less than 85 microvolts?
c) About how many of the 456 signals have strengths above 101 microvolts? (nearest integer)
d) About how many of the 456 signals have strengths between 90 and 100 microvolts?(nearest integer)
Explanation / Answer
Mean strength = 95 microvolts
Standard deviation of =5.7 microvolts
a) About what proportion of ASD's have a strength between 90 and 100 microvolts?
Pr(90 < X < 100) = Pr( X < 100 ; 95; 5.7) - Pr(X < 90; 95; 5.7) = (Z2 ) - (Z1)
where is the normal standard cumulative distribution function.
Z2 = (100 - 95)/5.7 = 0.8772
Z1 = (90 - 95)/5.7 = -0.8772
Pr(90 < X < 100) = = (Z2 ) - (Z1) = (0.8772) - (-0.8772) = 0.8098 - 0.1902 = 0.6196
b) About proportion of ASD's have strengths less than 85 microvolts?
Pr(X < 85) = Pr(X < 85; 95; 5.7)
Z = (85 - 95)/ 5.7 = -1.7544
Pr(X < 85) = Pr(X < 85; 95; 5.7) = Pr(Z < -1.7544) = 0.0397
(C) Pr(X > 101) = 1 - Pr(X < 101 ; 95; 5.7)
Z = (101 - 95)/5.7 = 1.0526
Pr(X > 101) = 1 - Pr(X < 101 ; 95; 5.7) = 1 - Pr(Z < 1.0526) = 1 - 0.8537 = 0.1463
signals have strengths above 101 microvolts out of 456 = 456 * 0.1463 = 66.71 or 67
(D) how many of the 456 signals have strengths between 90 and 100 microvolts?
= 456 * Pr(90< X < 100) = 456 * 0.6196 = 282.54 or 283
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