1. Given a random variable X with a normal distribution, with \\mu= 50 and \\sig

Question

1. Given a random variable X with a normal distribution, with mu= 50 and sigma= 10, find the probability that X assumes a value between 45 and 62.
2. For the previous problem, find the value of X which represents the 95th percentile.
3. The probability that a patient recovers from a rare blood disease is 0.4. If 100 people are known to have contracted the disease, what is the probability that less than 30 survive? (Use a normal approximation of a binomial distribution. Be sure to include a correction for continuity.)

Explanation / Answer

(1)

Mean, m = 50,

SD, S = 10

When x = 45, z-score = (45-50)/10 = -0.5

At this z-score, looking from the cumulative z-score, cumulative probability, p1 = 0.308

When x = 62, z-score = (62-50)/10 = 1.2

At this z-score, looking from the cumulative z-score, cumulative probability, p2 = 0.884

So, reqd. probability = p2-p1 = 0.884-0.308 = 0.576

(2)

At cumulative probability of 0.95, corresponding z-score is:

z = 1.6

X-value = m + z*S = 50 + 10*1.6 = 66

(3)

Given:

p = 0.4, n = 100

Sample mean = n*p = 40

Sample SD = (n*p*(1-p))0.5 = (100*0.4*0.6)0.5 = 4.89

At X = 30, z-score = (30-40)/4.89 = -2.044

At this z-score, cumulative probability = 0.020

Thus the required probability is:

P(X <= 30) = 0.02

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