Home page for accesible maths 2.3 Normal distribution 2.3.1 Normal distribution model 2.3.3 Calculating Normal Probabilities

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2.3.2 Standardizing with Z scores

Example 2.3.1

Table 2.1 shows the mean and standard deviation for total scores on the SAT and ACT. The distribution of SAT and ACT scores are both nearly normal. Suppose Ann scored 1800 on her SAT and Tom scored 24 on his ACT. Who performed better?

Answer. We use the standard deviation as a guide. Ann is 1 standard deviation above average on the SAT: $1500+300=1800$ . Tom is 0.6 standard deviations above the mean on the ACT: $21+0.6\times 5=24$ . In Figure LABEL:satActNormals, we can see that Ann tends to do better with respect to everyone else than Tom did, so her score was better.

	SAT	ACT
Mean	1500	21
SD	300	5

Table 2.1: Mean and standard deviation for the SAT and ACT.

Example 2.3.1 used a standardization technique called a Z score, a method most commonly employed for nearly normal observations but that may be used with any distribution. The Z score of an observation is defined as the number of standard deviations it falls above or below the mean. The Z score is also sometimes know as the standardized observation. If the observation is one standard deviation above the mean, its Z score is 1. If it is 1.5 standard deviations below the mean, then its Z score is -1.5. If $x$ is an observation from a distribution $N(\mu,\sigma^{2})$ , we define the Z score mathematically as

\displaystyle Z=\frac{x-\mu}{\sigma}

Using $\mu_{SAT}=1500$ , $\sigma_{SAT}=300$ , and $x_{Ann}=1800$ , we find Ann’s Z score:

\displaystyle Z_{Ann}=\frac{x_{Ann}-\mu_{SAT}}{\sigma_{SAT}}=\frac{1800-1500}{% 300}=1

The Z score The Z score of an observation is the number of standard deviations it falls above or below the mean. We compute the Z score for an observation $x$ that follows a distribution with mean $\mu$ and standard deviation $\sigma$ using $\displaystyle Z=\frac{x-\mu}{\sigma}$

Example 2.3.2

Use Tom’s ACT score, 24, along with the ACT mean and standard deviation to compute his Z score.

Answer. $Z_{Tom}=\frac{x_{Tom}-\mu_{ACT}}{\sigma_{ACT}}=\frac{24-21}{5}=0.6$ Observations above the mean always have positive Z scores while those below the mean have negative Z scores. If an observation is equal to the mean (e.g. SAT score of 1500), then the Z score is $0$ .

Example 2.3.3

Let $X$ represent a random variable from $N(\mu=3,\sigma^{2}=4)$ , and suppose we observe $x=5.19$ . (a) Find the Z score of $x$ . (b) Use the Z score to determine how many standard deviations above or below the mean $x$ falls.

Answer. (a) Its Z score is given by $Z=\frac{x-\mu}{\sigma}=\frac{5.19-3}{2}=2.19/2=1.095$ . (b) The observation $x$ is 1.095 standard deviations above the mean. We know it must be above the mean since $Z$ is positive.