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The solutions to quiz 3 for math 183, a college-level mathematics course taken in the fall of 2006. The solutions cover topics such as expected values of continuous random variables, binomial distribution, and confidence intervals. Students can use this document to check their understanding of these concepts and to study for exams.
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1a. The expected value of a continuous random variable is given by
E[Y ] =
yfY (y)dy
where the integral is over the domain of fY (y). So for part a we have
0
y · 3(1 − y)^2 dy
0
(3y^3 − 6 y^2 + 3y)dy
1b. Here we have
E[Y ] =
−∞
fY (y)dy
0
y ·
dy + 0 +
2
y ·
dy + 0
is given by integrating the pdf:
(^12) 3 y^2 dy
= y^3
1 (^12)
=
We now let X be the number of observations in this interval out of 15 samples. This is equivalent to flipping a biased coin with probability 78 of heads 15 times, and counting the number of heads. In other words, X is binomially distributed with n = 15 and p = 78. The expected value of X can be computed from the formula for the expected value of a binomial random variable:
E[X] = np = 15
3a. The “confidence” of an interval is the area over the corresponding region under the normal curve. So in part a, the confidence is:
ϕ(2.33) − ϕ(− 1 .64) ≈ 0. 99 − 0. 05 ≈ 0. 94
3b. Since the left endpoint is −∞, we just have the area to the left of the right endpoint.
ϕ(2.05) ≈ 0. 98
3c. Since y = y + 0 · σ/
n, we have
ϕ(0) − ϕ(− 1 .64) ≈ 0. 50 − 0. 05 ≈ 0. 45
X − z α 2
σ √ n
≤ p ≤ X + z α 2
σ √ n
≈ 1 − α
We don’t know σ, but since n is large
σ ≈
We wish to construct a 96% confidence interval, so α = 0.04 and
z α 2 = z 0. 02 ≈ 2. 05
Thus our confidence interval is [
z α 2 = z 0. 01 ≈ 2. 33
