Download Approximately Normally Distributed Discrete Random Variables

MAT132 Intro to Stats
Normal Approximation
Kåre S. Gjaldbæk
Approximately Normally Distributed Discrete
Random Variables
As we know, the normal distribution is a continuous probability distribution. So, when a discrete random variable approximately follows the normal
distribution, we must use our thinker.
The number of banjos owned by a Kentuckian is a discrete
random variable (no half banjos) which is approximately normally distributed
with a mean of 63 and a standard deviation of 14.
Problem 1.
(i) What is the probabilty that a randomly chosen Kentuckian owns 70
banjos?
(ii) What is the probabilty that a randomly chosen Kentuckian owns at
most 50 banjos?
(iii) What is the probabilty that a randomly chosen Kentuckian owns more
than 40 banjos?
(iv) What is the probabilty that a randomly chosen Kentuckian owns between 60 and 70 banjos?
(v) What is the middle 50% interval of banjos own by Kentuckians?
Draw the normal distribution on top of the probability histogram
of the discrete variable to see what's going on. We denote by X the discrete
variable, and by x the continuous normal variable.
Solution.
(i) We want to nd (approximate) P (X = 70), which we know translates
to the area of the bar of the histogram above 70. To approximate the area
of the bar above X = 70, we nd the area under the graph of the normal
distribution between x = 69.5 and x = 70.5. That is, nd
P (69.5 < x < 70.5).
First, we convert to the standard distribution. For x = 69.5, we have
z=
69.5 − 63
x−µ
=
= 0.46.
σ
14
z=
x−µ
70.5 − 63
=
= 0.54.
σ
14
For x = 70.5, we have
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MAT132 Intro to Stats
Normal Approximation
Kåre S. Gjaldbæk
So, using Table 3, we nd
P (69.5 < x < 70.5) = P (0.46 < z < 0.54)
= P (z < 0.54) − P (z < 0.46)
= 0.7054 − 0.6772
= 0.0282 = 2.82%
(ii) Now, it's pretty straight forward, the only caveat being to add ±0.5 to
adjust for continuity. To approximate P (X ≤ 50), we calculate P (x < 50.5).1
For x = 50.5, we have
z=
x−µ
50.5 − 63
=
= −0.89.
σ
14
So, using Table 3, we nd
P (X ≤ 50) ≈ P (x < 50.5)
= P (z < −0.89)
= 0.1867 = 18.67%
(iii)(v) Try it yourselves.
Approximating the Binomial Distribution
We probably (pun intended) love no discrete probability distribution more
than the binomial distribution. It just so happens that, under usual circumstances, the binomial distribution is approximately normally distributed
(remember how the histogram looks). If a binomial random experiment consists of n repeated trials, with succes probability p and q = 1 − p, then, as a
rule of thumb, if we have
np ≥ 5 and nq ≥ 5,
then the binomial distribution will approximately follow the normal distribution with the same mean and standard deviation as the binomial.
Note that if the problem was P (X < 50) (rather than P (X ≤ 50)), then the approximation would be P (x < 49.5).
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MAT132 Intro to Stats
Normal Approximation
Kåre S. Gjaldbæk
As I'm sure you're aware, before turning full time mathematician, I was a procient darts player. In fact, I hit the board in roughly 28%
of my attempts. One evening in a bar, my friend Todd says:
Problem 2.
I'll give you 10 times your money back if you hit the board at least
10 times in 25 attempts.
Should I take the bet?
Solution.
We want to nd
P (X ≥ 10) = P (X = 10) + P (X = 11) + · · · + P (X = 25),
which we could nd by calculating
n k
P (X = k) =
p (1 − p)n−k
k
25
=
0.28k (0.72)25−k ,
k
for k = 10, 11, 12, . . . , 25. We could shortcut this Herculean task by calculating for k = 0, 1, . . . , 9 and subtract from 1, but even that would take the
rest of the day.
Luckily, we see that
np = 25 · 0.28 = 7 and nq = 25 · 0.72 = 18.
Since both are greater than 5, we can justify approximating with the normal
distribution. The mean and standard deviation are
√
√
µ = np = 25 · 0.28,
σ = npq = 25 · 0.28 · 0.72 = 2.25.
To approximate P (X ≥ 10), we nd P (x > 9.5). First, we convert to
standard
x−µ
9.5 − 7
z=
=
= 1.11.
σ
2.25
Using Table 3, we nd
P (x > 9.5) = P (z > 1.11)
= P (z < −1.11)
= 0.1335 = 13.35%
So, since I'm getting 10 times my money back, I should take the bet.
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