5) A multiple choice test has 5 questions each of which has 4 possible answers.

Question

5) A multiple choice test has 5 questions each of which has 4 possible answers. is correct. If Fred, who forgot to study for the test, guesses on all questions, whato that he will answer exactly at least one of the questions correctly? Only one of the choices of the choices 6) Assume that the red blood cell counts of women are normally distributed with a and a standard deviation of 0.382. a) If one woman is selected at random, find the probability that: she has a red blood cell count below the normal range of 4.2 to 5.4. she has a red blood cell count above the normal range of 4.2 to 5.4 · b) Find the 95th percentile for the red blood cell counts of women. c) What percentage of women have red blood cell counts in the normal range? d) If 25 women are randomly selected, find the probability that the mean of their red blood cell count is I the normal range. If z is a standard normal variable, find the probability 7) The probability that z lies between -1.22 and 1.47 The graph depicts the standard normal distribution with mean 0 and standard deviation 1 8) Shaded area is 0.6215. Find the indicate z-score. Assume that X has a normal distribution, and find the indicated probability 9) The mean is = 97.0 and the standard deviation is 1.2. Find the probability that X is less than 100.0. The mean is = 1.52 and the standard deviation is Find the probability that X is greater than 0.9. 10) 0.61.

Explanation / Answer

5)
Probability that a randomly chosen answer is correct, p = 1/4 = 0.25

n=5

Consider an event where Fred answers none of the question correctly i.e. P(X = 0) = (1-p)^5 = (1-0.25)^5 = 0.2373

Required probability where at least one of the questions is answered correctly,
P(X>=1) = 1 - P(X=0)
= 1 - 0.2373
= 0.7627

6)
mean = 4.577
sigma = 0.382

a)
P(4.2 < X) = P((4.2 - 4.577)/0.382 < z) = P(-0.9869 < z) = 0.1618

P(X > 5.4) = P(z > (5.4 - 4.577)/0.382) = P(z > 2.1545) = 0.0156

b)
For 95%, z-value = 1.645

xbar = mean + z*sigma
xbar = 4.577 + 1.645*0.382 = 5.2054

c)
P(4.2 < X < 5.4) = P((4.2 - 4.577)/0.382 < z < (5.4 - 4.577)/0.382) = P(-0.9869 < z < 2.1545) = 0.8226

d)
P(4.2 < X < 5.4) = P((4.2 - 4.577)/(0.382/sqrt(25)) < z < (5.4 - 4.577)/(0.382/sqrt(25))) = P(-4.9 < z < 10) = 1

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