Question

      Suppose the force acting on a column that helps to support a building is a normally distributed random variable X with mean value 11.0 kips and standard deviation 1.25 kips. Compute the following probabilities by standardizing and then using a standard normal curve table from the Appendix Tables. (Round your answers to four decimal places.)

How do you standarize?

P(X less than or equal to 11)

P(X less than or equal to 13.5)

P(X greater than or equal to 4.75)

P(X in between 10 and 13 inclusive)

P(absolute value(X-11) less than or equal to 3)

Answer 1

Concepts and reason

A Z-score indicates the number of standard deviations an element is deviated from the mean. Z-scores may be positive or negative. The positive value indicates that the score is above the mean value. The negative value indicates that the score is below the mean value. The normal probability values can be determined with the help of Z-score.

Fundamentals

A Z-score can be calculated by using the following formula:

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

Here,

The value of the element is denoted by x.

The mean of the population is denoted by $\mu$ .

The standard deviation of the population is denoted by $\sigma$ .

(a)

If a continuous random variable X is said to follow normal distribution with mean $\mu$ and standard deviation $\sigma$ is said to follow normal distribution.

Usually standard normal variate is denoted by Z.

Hence, standardize the random variable x by using

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

(b)

From the given information the mean and standard deviation of the random variable X is 11 and 1.25.

The probability that the random variable X is less than or equal to 11 is calculated as follows:

$\begin{array}{c}\\P\left( {X \le 11} \right) = P\left( {\frac{{X - \mu }}{\sigma } \le \frac{{11 - 11.0}}{{1.25}}} \right)\\\\ = P\left( {Z \le \frac{0}{{1.25}}} \right)\\\\ = P\left( {Z \le 0} \right)\\\\ = 0.5000 & & \left( {{\rm{From}}\,\,{\rm{standard}}\,{\rm{Normal}}\,{\rm{curve}}\,{\rm{tables}}} \right)\\\end{array}$

(c)

From the given information the mean and standard deviation of the random variable X is 11 and 1.25.

The probability that the random variable X is less than or equal to 13.5 is calculated as follows:

$\begin{array}{c}\\P\left( {X \le 13.5} \right) = P\left( {\frac{{X - \mu }}{\sigma } \le \frac{{13.5 - 11.0}}{{1.25}}} \right)\\\\ = P\left( {Z \le \frac{{2.5}}{{1.25}}} \right)\\\\ = P\left( {Z \le 2} \right)\\\\ = 0.9772 & & \left( {{\rm{From}}\,\,{\rm{standard}}\,{\rm{Normal}}\,{\rm{curve}}\,{\rm{tables}}} \right)\\\end{array}$

(d)

From the given information the mean and standard deviation of the random variable X is 11 and 1.25.

The probability that the random variable X is greater than or equal to 4.75 is calculated as follows:

$\begin{array}{c}\\P\left( {X \ge 4.75} \right) = 1 - P\left( {X < 4.75} \right)\\\\ = 1 - P\left( {\frac{{X - \mu }}{\sigma } < \frac{{4.75 - 11.0}}{{1.25}}} \right)\\\\ = 1 - P\left( {Z < \frac{{ - 6.25}}{{1.25}}} \right)\\\\ = 1 - P\left( {Z < - 5} \right)\\\\ = 1 - 0.0000 & & \left( {{\rm{From}}\,\,{\rm{standard}}\,{\rm{Normal}}\,{\rm{curve}}\,{\rm{tables}}} \right)\\\\ = 0.9999\\\end{array}$

(e)

From the given information the mean and standard deviation of the random variable X is 11 and 1.25.

The probability that the random variable X lies between 10 and 13 both inclusive is calculated as follows:

$\begin{array}{c}\\P\left( {10 \le X \le 13} \right) = P\left( {X \le 13} \right) - P\left( {X \le 10} \right)\\\\ = P\left( {\frac{{X - \mu }}{\sigma } \le \frac{{13 - 11.0}}{{1.25}}} \right) - P\left( {\frac{{X - \mu }}{\sigma } \le \frac{{10 - 11.0}}{{1.25}}} \right)\\\\ = P\left( {Z \le 1.6} \right) - P\left( {Z \le - 0.8} \right)\\\\ = 0.9452 - 0.2118\\\\ = 0.7334 & \left( {{\rm{From}}\,\,{\rm{standard}}\,{\rm{Normal}}\,{\rm{curve}}\,{\rm{tables}}} \right)\\\end{array}$

(f)

From the given information the mean and standard deviation of the random variable X is 11 and 1.25.

The required probability is calculated as follows:

$\begin{array}{c}\\P\left( {\left| {X - 11} \right| < 3} \right) = P\left( {\left| Z \right| < \frac{3}{{1.25}}} \right)\\\\ = P\left( {\left| Z \right| < 2.4} \right)\\\\ = P\left( { - 2.4 < Z < 2.4} \right)\\\\ = P\left( {Z < 2.4} \right) - P\left( {Z < - 2.4} \right)\\\\ = 0.9918 - 0.0082\\\\ = 0.9836 & \left( {{\rm{From}}\,\,{\rm{standard}}\,{\rm{Normal}}\,{\rm{curve}}\,{\rm{tables}}} \right)\\\end{array}$

Ans: Part a

Random variable X is standardize by using,

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

Part b

Probability that the random variable X is less than or equal to 11 is 0.5000.

Part c

Probability that the random variable X is less than or equal to 13.5 is 0.9772.

Part d

Probability that the random variable X is greater than or equal to 4.75 is 0.9999.

Part e

Probability that the random variable X lies between 10 and 13 both inclusive is 0.7334.

Part f

Probability that the absolute value (X-11) less than or equal to 3 is 0.9836.