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z-Scores Solutions

• Therefore:
  • $\underline{\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{𝜇}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{ }\mathbf{ }}$
  • $\underline{\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{𝜎}\mathbf{=}\mathbf{1}\mathbf{ }\mathbf{ }\mathbf{ }}$
• This type of distribution is a    density curve    with an area    always equal to one square unit   , or in terms of a proportion,    100%   .
• Because the mean is zero,    50% of the data is below the mean     (and 50% is above the mean).
• Any normally distributed data set can be standardized using the    z-score formula   : $z=\frac{X–\mu }{\sigma }$
• A z-score is a measure of how many standard deviations, 𝜎,    a data point is from the population mean   , 𝜇.

• The area can be written as a    proportion or percentage    from the z-table.

• The    top row    and    left column    correspond with the z-score.

• When analyzing z-values using the z-table:

  • Ask yourself: What proportion of the data    is less than X   ?

  • Write answers using the math shorthand: $\underline{\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{P}\mathbf{(}\mathbf{data}\mathbf{<}\mathit{X}\mathbf{)}\mathbf{ }\mathbf{ }\mathbf{ }}$.

Example 1

Standardize the normal distribution.

$$\begin{gathered} z=\frac{X–\mu }{\sigma } \\ \begin{array}{ccc}z=\frac{100–100}{5.5}=0& & z=\frac{83.5–100}{5.5}=–3\end{array} \end{gathered}$$

$\begin{array}{ccc}z=\frac{89–100}{5.5}=–2& & z=\frac{94.5–100}{5.5}=–1\\ & & \\ z=\frac{105.5–100}{5.5}=1& & z=\frac{111–100}{5.5}=2\\ & & \\ z=\frac{116.5–100}{5.5}=3& & \end{array}$

Example 2

A high school track coach calculated the post-season mile times to be $\mu =6.417, \sigma =0.333$. What proportion of runners on the track team ran a mile in less than 6:30 minutes?

Example 3

The area for a standard normal distribution is 0.0495 with a mean of 7.5 and a standard deviation of 1.5. Determine the raw data value. Explain.