need help with 3 please 3. The percentage of people in a population with a certa

Question

need help with 3 please

3. The percentage of people in a population with a certain ailment (Ailment A) is 12%. a. If you select a sample of 10 people from this population, what is the probability that at most two of them will have Ailment A? b. What is the probability that at least 3 of them would have this ailment ? c. If you select a sample of 100 people, what is the probability that less than 10 will have ailment A? Use the normal approximation technique. d. What is the probability, in your sample of 100, that at least 20 will have Ailment A? Again, use the normal approximation to the binomial as shown in class.

Explanation / Answer

Solution:

Given that p = 0.12. Using binomial distribution formula,

P (X = x) = nCx p^x (1 – p)^n-x

a. P (X ? 2) = P (X = 0) + P (X = 1) + P (X = 2)

                    = 10C0 (0.12)^0 (0.88)^10 + 10C1 (0.12)^1 (0.88)^9 + 10C2 (0.12)^2 (0.88)^8

                   = 0.2785 + 0.3798 + 0.2330

                   = 0.8913

b. P (X ? 3) = 1 – P (X < 3)

                    = 1 – [P (X = 0) + P (X = 1) + P (X = 2)]

                   = 1 – [0.2785 + 0.3798 + 0.2330]

                   = 0.1087

c. Mean, µ = np = 100*0.12 = 12, Standard deviation, ? = ?np (1 – p) = ?100*0.12*0.88 = 3.25

Using continuity correction, we have

P (X < 10) = P (X < 9.5)

The respective Z-score with X = 9.5 is

Z = (X - µ)/?

Z = (9.5 – 12)/3.25

Z = -0.77

Using Z-tables, the probability is

P [Z < -0.77] = 0.2207

d. Using continuity correction, we have

P (X ? 20) = P (X > 19.5)

The respective Z-score with X = 19.5 is

Z = (X - µ)/?

Z = (19.5 – 12)/3.25

Z = 2.31

Using Z-tables, the probability is

P [Z > 2.31] = 1 – 0.9896

                     = 0.0104

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