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Name: AMS 7 Homework Quiz 2 Wednesday, October 23, 2013 1. A sample of the blood of 100 people found the following blood types: Rh Type O Positive 39 Negative 6 45 Group A B AB 35 8 4 5 2 1 40 10 5 86 14 100 (a) (2.5 points) If one person is randomly selected, find the probability of getting someone who is group B or type Rh-. P (B or Rh−) = P (B) + P (Rh−) − P (B and Rh−) 14 2 22 10 + − = = 0.22 = 100 100 100 100 (b) (2.5 points) Given that we select a random person who is Rh-, what is the probability that they are group B? P (B|Rh−) = P (B and Rh−) 2/100 2 = = = 0.14 P (Rh−) 14/100 14 (c) (2 points) Suppose that people who are both group B and Rh- visit the blood clinic at the rate of three per week. Using the Poisson distribution, whamt would be the mean and standard deviation of the number of people who are both group B and Rh- who visit the clinic in three weeks? µ=3·3=9 √ √ σ= µ= 9=3 2. A leatherback turtle has made a nest with 110 eggs. Suppose each egg is independent and has a probability of 0.85 of hatching. (a) (2 points) What is the mean and standard deviation of the number of eggs that hatch? µ = n · p = 110 · 0.85 = 93.5 p √ σ = np(1 − p) = 110 · 0.85 · 0.15 = 3.745 (b) (2 points) Would it be unusual for 100 eggs to hatch? Why or why not? µ + 2σ = 93.5 + 2 · 3.745 = 100.99 So, NO, it is not unusual for 100 eggs to hatch. 3. Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15. (a) (3 points) Find the probability that a randomly selected adult gads an IQ that is less than 110. X −µ 110 − µ 110 − 100 P (X < 110) = P < =P Z< σ σ 15 10 = P (Z < ) = P (Z < 0.67) = 0.7486 15 (b) (3 points) Find the probability that a randomly selected adult has an IQ between 95 and 115. X −µ 115 − µ 95 − µ < < P (95 < X < 115) = P σ σ σ 95 − 100 115 − 100 −5 15 =P <Z< = P( <Z< ) 15 15 15 15 = P (Z < 1) − P (Z < −0.33) = 0.8413 − 0.3707 = 0.4706 (c) (3 points) Find P20 , which is the IQ score separating the bottom 20% from the top 80%. We want to find P (X < P20 ) = 0.20 We know that P (Z < −0.845) = 0.20 ⇒ P (Zσ + µ < −0.845σ + µ) = 0.20 (∗∗) ⇒ P (X < −0.845 · 15 + 100) = 0.20 ⇒ P (X < 87.325) = 0.20 So P20 = 87.325 (∗∗) Z = X −µ ⇒ X = Zσ + µ σ