Solve this problem using substitution
1 answer:
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(a) For any probability distribution, the total probability must be 1. That is, the area under the probability density curve must be equal to 1.
The empirical rule for normal distributions says
• approximately 68% of the distribution lies within 1 standard deviation of the mean
• approx. 95% lies within 2 s.d. of the mean
• approx. 99.7% lies within 3 s.d. of the mean
In this case, with mean 3500 and s.d. 470, this translates to
• Pr(3500 - 470 < X < 3500 + 470) = Pr(3030 < X < 3970) ≈ 0.68
• Pr(3500 - 2*470 < X < 3500 + 2*470) = Pr(2560 < X < 4440) ≈ 0.95
• Pr(3500 - 3*470 < X < 3500 + 3*470) = Pr(2090 < X < 4910) ≈ 0.997
Continuous probability distributions also have the property that
Pr(a < X < b) = Pr(a < X < c) + Pr(c < X < b)
if a < c < b.
Combining all these properties, we can find the probabilities for each of the 8 regions in the graph to be (from left to right)
• Pr(-∞ < X < 2090) ≈ (1 - 0.997)/2 ≈ 0.0015
• Pr(2090 < X < 2560) ≈ (1 - 0.95 - 2*0.0015)/2 ≈ 0.0235
• Pr(2560 < X < 3030) ≈ (1 - 0.68 - 2*0.0235 - 2*0.0015)/2 ≈ 0.135
• Pr(3030 < X < 3500) ≈ 0.68/2 ≈ 0.34
and since the distribution is symmetric about its mean, we already know the remaining probabilities,
• Pr(3500 < X < 3970) ≈ 0.34
• Pr(3970 < X < 4440) ≈ 0.135
• Pr(4440 < X< 4910) ≈ 0.0235
• Pr(4910 < X < ∞) ≈ 0.0015
(b) Per the rule, 99.7% of babies would weight between 2090 and 4910 grams.
(c) The proportion of babies weighing less than 3030 grams is the sum of the proportions of babies weighing less than 2090, between 2090 and 2560, and between 2560 and 3030 grams. So
Pr(X < 3030) = Pr(-∞ < X < 2090) + Pr(2090 < X < 2560) + P(2560 < X < 3030)
Pr(X < 3030) ≈ 0.0015 + 0.0235 + 0.135
Pr(X < 3030) ≈ 0.16 = 16%
(d) Similarly,
Pr(X > 2560) = Pr(2560 < X < 3030) + Pr(3030 < X < 3500) + … + Pr(4910 < X < ∞)
Pr(X > 2560) ≈ 0.135 + 0.34 + 0.34 + 0.135 + 0.0235 + 0.0015
Pr(X > 2560) ≈ 0.975 = 97.5%
