+
+ When a normal distribution gets translated and/or scaled, the result is another normal distribution.
+ It will often be useful to standardize a normal distribution X = \Norm(\mu, \sigma^2) as follows:
+
+ Z = \frac{X - \mu}{\sigma} = \Norm(0, 1).
+
+ Given an X-value a, the corresponding value of Z is called a z-score, and is sometimes denoted a^*
+
+
+
+
+
+ Suppose we flip a fair coin 100 times and let S_{100} be the number of heads.
+ In order to calculate, for example, \Pr(40 \leq S_{100} \leq 60) using the binomial distribution probability formula, we would need to calculate:
+
+ \Pr(40 \leq S_{100} \leq 60) = \sum_{k = 40}^{60} {100 \choose k} (0.5)^k (1 - 0.5)^{100 - k}.
+
+ This could be done using a good calculator.
+ If we increased the number of flips to 1000, or 10000, etc., even a powerful calculator might struggle with the computation.
+ However, if we're willing to accept an approximation, then gives a (mostly) straightforward path:
+
+ \E(S_{100}) \amp = np = (100)(0.5) = 50
+ \Var(S_{100}) \amp = np(1-p) = (100)(0.5)(0.5) = 25
+ \Rightarrow \quad S_{100} \amp\approx \Norm(50, 25)
+ Z \amp = \frac{S_{100} - 50}{\sqrt{25}}
+ \Pr(40 \leq S_{100} \leq 60) \amp\approx \Pr\left(\frac{40 - 50}{5} \leq Z \leq \frac{60 - 50}{5}\right)
+ \amp= \Pr\left(-2 \leq Z \leq 2\right)
+
+ We now reach a potential impass.
+
+
+
+
+
+ The random variable Z is the standard normal distribution, with pdf:
+
+ \phi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}
+
+ To calculate probabilities like \Pr(-2 \leq Z \leq 2), we should evaluate the integral:
+
+ \Pr(-2 \leq Z \leq 2) = \int_{-2}^2 \phi(x)\ dx = \int_{-2}^2 \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\ dx
+
+ Alternatively, we could use the cdf:
+
+ \Phi(x) = \int_{-\infty}^x \phi(t)\ dt = \int_{-\infty}^x \frac{1}{\sqrt{2\pi}} e^{-t^2/2}\ dt
+
+ If we had a nice formula for \Phi(x), then we could complete the probability calculation:
+
+ \Pr(-2\leq Z \leq 2) = \Phi(2) - \Phi(-2).
+
+ Unfortunately, we don't have a nice formula for \Phi(x).
+ In fact, there is provably no "nice" formula that could be written for \Phi(x) (where "nice" has some technical mathematical meaning that we won't discuss here).
+
+
+
+ The true power of comes to the rescue: it doesn't matter what type of distribution the individual X_i's have, and it doesn't matter what type of distributions S_n and A_n have.
+ The random variable Z is \Norm(0, 1).
+ Since it's the same particular distribution all the time, and since we're only aiming at an approximation anyway, we can calculate many estimations of probabilities like \Pr(a \leq Z \leq b) ahead of time and just save the results to look up later.
+
+
+
+ A table of precalculated \Phi(z) values can be found in .
+
+
+
+
+
+ Continuing the previous example, and having looked up the values \Phi(2) \approx 0.9772 and \Phi(-2) \approx 0.0228, we complete our estimation:
+
+ \Pr(40 \leq S_{100} \leq 60) \amp\approx \Pr\left(\frac{40 - 50}{5} \leq Z \leq \frac{60 - 50}{5}\right)
+ \amp= \Pr\left(-2 \leq Z \leq 2\right)
+ \amp \approx \Phi(2) - \Phi(-2)
+ \amp \approx 0.9772 - 0.0228
+ \amp = 0.9544.
+
+
+
+
+
+
+ The true value is approximately 0.965, so we underestimated the probability a bit.
+ It turns out this will consistently be the case when using a normal approximation to a discrete distribution, such as the binomial distribution.
+ The range of values 40 \leq S_{100} \leq 60 is a list of 21 discrete values.
+ However, the continuous interval [40, 60] has a length of only 20.
+ So calculating the z-scores from the values 40 and 60 has the effect of shortening the interval length by 1.
+ To mitigate this, we can make a continuity correction by extending the values out by 0.5 in each direction:
+
+ \Pr(40 \leq S_{100} \leq 60) \amp\approx \Pr\left(\frac{39.5 - 50}{5} \leq Z \leq \frac{60.5 - 50}{5}\right)
+ \amp= \Pr\left(-2.1 \leq Z \leq 2.1\right)
+ \amp \approx \Phi(2.1) - \Phi(-2.1)
+ \amp \approx 0.9821 - 0.0179
+ \amp = 0.9642.
+
+ This is a much more accurate estimate.
+
+
+
+
+
+
+ Suppose a coin has probability 0.4 of coming up heads, and we flip the coin 100 times.
+ Let S be the number of heads.
+ Estimate the probability that 34 \leq S \leq 44.
+
+
+
+
+
+ \Pr(34 \leq S \leq 44) \approx 0.8212 - 0.0918 = 0.7294.
+
+
+
+
+
+
+
+ Suppose a fair die is rolled 100 times, and let m be the average value of the rolls.
+ Estimate the probability that 3.45 \leq m \leq 3.55.
+
+
+
+
+
+ \Pr(3.45 \leq m \leq 3.55) \approx 0.6141 - 0.3859 = 0.2282.
+
+
+
+
+
+
+
+ The heights of men in the US have a mean of 69 in and a variance of about 9 in^2, and the heights of women in the US have a mean of 63.5 in with a variance of 6.25 in^2.
+
+
+
+
+
- Suppose a coin has probability 0.4 of coming up heads, and we flip the coin 100 times.
- Let S be the number of heads.
- Estimate the probability that 34 \leq S \leq 44.
+ Suppose the heights of 30 men are sampled, and a sample mean m is taken.
+ Find the expected value and variance of m.
-
-
+
+
+ \E(m) = 69 and \Var(m) = 0.3.
+
+
+
+
+
+
- Suppose a fair die is rolled 100 times, and let m be the average value of the rolls.
- Estimate the probability that 3.45 \leq m \leq 3.55.
+ Estimate the probability that m \geq 69.5.
-
-
-
+
- The heights of men in the US have a mean of 69 in and a variance of about 9 in^2, and the heights of women in the US have a mean of 63.5 in with a variance of 6.25 in^2.
+ \Pr(m \geq 69.5) \approx 1 - 0.8186 = 0.1814.
-
+
+
-
-
-
- Suppose the heights of 30 men are sampled, and a sample mean m is taken.
- Find the expected value and variance of m.
-
-
-
+
+
+
+ If the sampled group was women, estimate the probability that m \geq 64.
+
+
-
-
-
-
- Estimate the probability that m \geq 69.5.
-
-
-
-
-
-
-
-
- What if the sampled group was women?
-
-
-
-
-
-