Solution to Practice Quiz 8
The full quiz is here. The
answers appear below. Comments, which are not part of the answers, are italicized.
Time: 20
minutes. This quiz is open book, open notes.
1. Identify the statements that must be
false and state why. Explain why the remaining statements can be true.
- A random variable X has a mean of 3, a standard deviation
of 2, and the probability that X is 9 or greater is 10%.
- A random variable X is lognormally distributed with a
mean of 20 and a standard deviation of 20. The probability that X is
less than zero is 1%.
- Summary statistics for a batch of 30 data include a mean
of 101.4, a standard deviation of 23.3, a fifth percentile of 13.0, and a
95th percentile of 179.9.
- Every distribution has a median.
- Every distribution has a mean.
- The formula, "if today is a Tuesday, Thursday, or
Saturday, then compute
; otherwise compute
" is an estimator of the standard deviation of a statistical sample X1,
..., Xn, provided n > 1.
a. Standardizing gives 9 = (9-3)/2 = 3
sds above the mean. Chebyshev's inequality states that the probability of
exceeding a 3 sd difference from the mean is no greater than 1/32 =
1/9 = 111/9%. It appears, then, that (a) could be true.
To verify this, we would need to construct a distribution with a mean of 3,
an sd of 2, and a 10% probability of exceeding 9. A discrete distribution
with 90% probability of the outcome 21/3 and a 10%
probability of the outcome 9 has these properties. This is easy to check,
because shifting the location by -21/3 and scaling by 3/20
gives a B(1, 0.10) distribution with mean 0.10 and variance 0.10 * (1 - 0.10) =
0.09. Therefore the original distribution has mean 20/3 * 0.10 + 21/3
= 3 and variance (20/3)2 * 0.09 = 4, so its sd is 2, as desired.
b. This is false, because all
lognormal distributions have non-negative support (they can never produce
negative values).
c. Standardizing, we see that the fifth
percentile is (13 - 101.4)/23.3 = 3.8 sds below the mean and the 95th percentile
is (189.9 - 101.4)/23.3 = 3.4 sds above the mean. Therefore, the
probability of being at least 3.4 sds away from the mean is at least 5 +
(100-95) = 10 percent. However, Chebyshev's Inequality states that this
probability cannot exceed 1/3.42 = 8.8 percent, which is a
contradiction. We conclude this statement is false.
d. True, because every
distribution has a CDF which rises from 0 to 100% and therefore must cross 50%
at some point (or points). That crossing point (or points) is the median,
by definition.
e. False. There is no reason
to suppose the mean should exist and it's easy to construct counterexamples,
such as the Cauchy distribution, that have no mean (it is effectively infinite).
f. False. This formula
depends on more than the sample results. It also depends importantly on a
completely extraneous value (the day of the week). Thus it does not fit
the definition of an estimator.
2. A statistical sample of 10 values is
obtained from a lognormal distribution with mean 30 and standard deviation
50. Does the skewness of the sample have to be positive? Why or why
not?
No. The sample is a random outcome. It could,
by chance alone, have a negative skewness. Indeed, any batch of
positive numbers has some chance, no matter how small, of resulting from a
statistical sampling of a lognormal distribution.
Scoring: The passing score is 95.
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