In 2007, Bob Jones University ended its fall semester a week early because of a whooping cough outbreak; 158 students were isolated with suspected infection and another 1200 given antibiotics as a precaution.14 Authorities react strongly to whooping cough outbreaks because the disease is so contagious. Because the effect of childhood vaccination often wears off by late adolescence, treat the Bob Jones students as if they were unvaccinated. It appears that approximately 1400 students were exposed. What is the probability that at least 75% of these students would develop infections if not treated

If not treated, the probability of at least 75 percent of 1400 students being infected is approximately equal to 1.

Step-by-step explanation:

Let Y be the no. of infections

The number of students who have been exposed to whooping cough is as follows:

n = 1400

n = 1400  and p = 0.8

nm = 1400 x 0.8

nm = 1120

n(1-m) = 1400 x (1-0.8)

n(1-m) = 280

Here,

Both nm and n(1-m) are greater than or equal to 10. So, we use normal distribution.

The mean and standard deviation of number of new infections is:

u = np

u =1400 x 0.8

u = 1120 = mean

SD = Standard Deviation  

SD =

SD = 14.966

The number of infections is distributed in an approximately normal way

N(1120,14.966)

75% of 1400 means 0.75  x 1400 = 1050

So,

Probability required.

P (Y 1050) = P (Y 1050 - 0.5)    (Continuity Correction)

P (Y 1050) = P()

P (Y 1050) =  P(Z -4.71)

P (Y 1050) = P(Z 4.71)

Using the z table:

P (Y 1050) ≈ 1

Hence,

If not treated, the probability of at least 75 percent of 1400 students being infected is approximately equal to 1.

2^y = ln 5, or   y(ln 2) = ln 5.   solving for y,   y = (ln 5) / (ln 2).   y=5

step-by-step explanation:

∠cbe and ∠deb are same-side interior angles.

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