2. Childhood lead poisoning is a public health concern in the United States. In
ID: 2921338 • Letter: 2
Question
2. Childhood lead poisoning is a public health concern in the United States. In a certain population, 1 child in 8 has a high blood lead level. In a randomly chosen group of 16 children from the population what is the probability that (a) 1 has high blood lead? (b) 3 or more have high blood lead? (c) How many children are not expected to have high blood lead levels? 3. Given that X is a random variable having a Poisson distribution, compute the following: (a) P(X = 8) when = 6 (b) P(X 2) when = 3.5 4. Every week the average number of wrong-number phone calls received by a certain-mail order house is seven. What is the probability that they will receive (a) two wrong calls tomorrow? (b) less than three wrong calls tomorrow? 5. Problem 4.1.2 in textbook 6. The thickness of photoresist applied to wafers in semiconductor manufacturing at a particular location on the wafer is uniformly distributed between 0.2050 and 0.2150 micrometers. Determine the following: (a) Cumulative distribution function (c.d.f.) of photoresist thickness (b) Probability that a wafer exceeds 0.2125 micrometers in photoresist thickness (c) Thickness such that the probability that a wafer exceeds it is 0.1Explanation / Answer
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 16 * 0.125
= 2
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 16 * 0.125 * 0.875
= 1.75
III.
standard deviation = sqrt( variance ) = sqrt(1.75
=1.32288
a.
P( X = 1 ) = ( 16 1 ) * ( 0.125^1) * ( 1 - 0.125 )^15
= 0.26987
b.
P( X < 3) = P(X=2) + P(X=1) + P(X=0)
= ( 16 2 ) * 0.125^2 * ( 1- 0.125 ) ^14 + ( 16 1 ) * 0.125^1 * ( 1- 0.125 ) ^15 + ( 16 0 ) * 0.125^0 * ( 1- 0.125 ) ^16
= 0.67708
P( X > = 3 ) = 1 - P( X < 3) = 0.32292
c.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 16 * 0.875
= 14
around 14 children are not expected to have high blood lead levels
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