91Ó°ÊÓ

The term "probability" simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

Given: Normal distribution

\(\begin{array}{l}{\rm{\mu = 104}}\\{\rm{\sigma = 5}}\end{array}\)

(a) The standardized score is the value \({\rm{x}}\)decreased by the mean and then divided by the standard deviation.

\(\begin{array}{c}{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}\\{\rm{ = }}\frac{{{\rm{105 - 104}}}}{{\rm{5}}}\\{\rm{\gg 0}}{\rm{.20}}\end{array}\)

Determine the corresponding probability using table \({\rm{A}}{\rm{.3}}\):

\(\begin{array}{l}P(X < 105) = P(Z < 0.20) = 0.5793\\P(X \le 105) = P(Z < 0.20) = 0.5793\end{array}\)

The probability of a continuous random variable being equal to a specific value is always zero (Note: you can notice this by the probabilities \({\rm{P(X < 105)}}\)and \(P(X \le \)\({\rm{105}}\)) that are equal):

\({\rm{P(X = 105) = 0}}\)

b)

\({\rm{x}}\)is \({\rm{1}}\) standard deviation from the mean:

\({\rm{x = \mu \pm \sigma }}\)

The standardized score is the value \({\rm{x}}\)decreased by the mean and then divided by the standard deviation.

\(\begin{array}{c}{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}\\{\rm{ = }}\frac{{{\rm{\mu \pm \sigma - \mu }}}}{{\rm{\sigma }}}\\{\rm{ = }}\frac{{{\rm{ \pm \sigma }}}}{{\rm{\sigma }}}\\{\rm{ = \pm 1}}{\rm{.00}}\end{array}\)

Determine the corresponding probability using table \({\rm{A}}{\rm{.3}}\):

\(\begin{array}{c}{\rm{P(|X| > \mu \pm \sigma ) = P(Z < - 1 or Z > 1)}}\\{\rm{ = 2P(Z < - 1) = 2(0}}{\rm{.1587)}}\\{\rm{ = 0}}{\rm{.3174}}\end{array}\)

Note: the probability is not dependent on the values of \({\rm{\mu }}\)and\({\rm{\sigma }}\).

c)

The most extreme \({\rm{0}}{\rm{.1\% }}\)of chloride concentration values are the lowest \({\rm{0}}{\rm{.05\% }}\)of the chloride values and the highest \({\rm{0}}{\rm{.05\% }}\) (lowest\({\rm{99}}{\rm{.95\% }}\)) of the chloride values (using that the normal distribution is symmetric about the mean).

Determine the \({\rm{z}}\)-score corresponding to a probability of \({\rm{0}}{\rm{.05\% }}\) (0.0005), and \({\rm{99}}{\rm{.95\% (0}}{\rm{.9995)}}\) using table A.3:

\({\rm{z = \pm 3}}{\rm{.29}}\)Note: there are multiple z-score with probability \({\rm{0}}{\rm{.0005/0}}{\rm{.9995}}\)in table \({\rm{A}}{\rm{.3}}\), thus we used technology to narrow the score down further (you could also take the average score, which would be\({\rm{ \pm 3}}{\rm{.295}}\)).

The corresponding value is the mean increased by the product of the \({\rm{z}}\)-score and the standard deviation:

\(\begin{array}{c}{\rm{x = \mu + z\sigma = 104 - 3}}{\rm{.29(5)}}\\{\rm{ = 87}}{\rm{.55x = \mu - z\sigma }}\\{\rm{ = 104 + 3}}{\rm{.29(5)}}\\{\rm{ = 120}}{\rm{.45}}\end{array}\)

Thus, the most extreme \({\rm{0}}{\rm{.1\% }}\)of chloride concentration values are below \({\rm{87}}{\rm{.55mmol/L}}\)and above\({\rm{120}}{\rm{.45mmol/L}}\).