Download (1 ) (5)(0.2)(0.8) 0.8 0.894 np p

K12MATHAPSTATS
Statistics Answers to Flexbook Section 4.5
1. It is easiest to use the calculator’s binomial pdf function to obtain each of the
probabilities. Press 2nd and then VARS, and scroll down to Option A (binompdf). For
p(0), enter n = 5, p = 0.2, and x = 0. Scroll down to Paste and press Enter. Press Enter
again, and the answer is 0.4096.
Repeat the procedure for x = 1, x = 2, x = 3, and x = 4. Here are the results, presented
in tabular form:
X
0
1
2
3
4
p(X) 0.4096 0.4096 0.1536 0.0256 0.0016
2. Use the same procedure as you did in problem 1, with n = 5 and p = 0.2 Use x values
of 0, 1, 2, 3, 4, and 5. Here is the probability distribution:
X
0
1
2
3
4
5
p(X) 0.328 0.4096 0.2048 0.0512 0.0064 0.0003
We calculate the mean µ by the formula   np , so   np  (5)(0.2)  1 .
The standard deviation is   np(1  p)  (5)(0.2)(0.8)  0.8  0.894 .
3. We have a binomial experiment with n = 5 and p = 0.1.
5!
(0.1)0 (1  0.1)5  (1)(1)(0.9)5  0.59 . This means that the
0!(5  0)!
probability that none of the five people chosen will develop antibodies is about .59.
3.a. P( X  0) 
3.b. The probability that at least one will develop antibodies is p(1) + p(2) + p(3) + p(4) +
p(5). Recognize that this is the complement of the question raised in part (a), so to find
the probability, just subtract the probability from part (a) from 1. Use the Complement
Rule to obtain p(at least 1) = 1 – P(0) = 1 – 0.59 = 0.41.
3.c. “Fewer than 3” means 0 or 1 or 2. We can use the calculator’s binomial cumulative
distribution function to determine this. Choose Option B (binomcdf), entering n = 5, p =
0.1, and x-value of 2. The answer is 0.991.
d. “More than 4” means “5.” We use Option A (binompdf), entering n = 5, p = 0.1, and x
= 5. The answer is 0.00001, a very small probability.
4. For this problem, we have n = 24 and p = 0.5.
Saylor URL: www.saylor.org/K12MATHAPSTATS/#4.5
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4.a. The expected value is another name for the mean. Use E(X) =   np as follows:
  np  (24)(0.5)  12 .
4.b. Calculate the standard deviation as   np(1  p)  (24)(.5)(.5)  6  2.45 .
4.c. P(X ≥ 18) = 1- P(X ≤ 17). We choose the Rule of the Complement so that we can
use the calculator for this computation. Use Option B (binomcdf), with n = 24, p = 0.5,
and x = 17. This will calculate the probability that x is less than or equal to 17, which is
the complement of the probability of interest. The calculator value is 0.989, so 1- P(X ≤
17) = 1 – 0.989 = 0.011.
Saylor URL: www.saylor.org/K12MATHAPSTATS/#4.5
The Saylor Foundation
Saylor.org
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