Question

6.5 Given a normal distribution with ? = 100 and ? =10
what is the probability that
a. x>75?
b. x<70?
c.x<80 or x>110?
d. 80% of the values are between what two X values (symmetrically
distributed around the mean)?

Answer 1

Concepts and reason

A z-score tells us, how many standard deviations and in which direction an observation falls away from the mean. 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.

For determining the z-score for an observation, one must know the mean and standard deviation of the distribution.

Fundamentals

The z-score can be determined as follows:

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

Since the total probability is always equal to 1, the probability of at least case can be solved as follows:

$\begin{array}{c}\\P\left( {X \ge x} \right) = 1 - P\left( {X < x} \right)\\\\ = 1 - P\left( {\frac{{X - \mu }}{\sigma } < \frac{{x - \mu }}{\sigma }} \right)\\\\ = 1 - P\left( {z < \frac{{x - \mu }}{\sigma }} \right)\\\end{array}$

Where,

$\mu$ is the mean of the distribution.

$\sigma$ is the variance of the distribution.

$x$ is the required value.

(a)

Let X be the random variable follows normal distribution with mean 100 and standard deviation 10.

$\mu = 100{\rm{ and }}\sigma = {\rm{10}}$

The probability of X greater than 75 is,

$\begin{array}{c}\\P\left( {X > 75} \right) = 1 - P\left( {X \le 75} \right)\\\\ = 1 - P\left( {\frac{{X - \mu }}{\sigma } \le \frac{{75 - 100}}{{10}}} \right)\\\\ = 1 - P\left( {Z \le - 2.5} \right)\\\\ = 1 - 0.0062{\rm{ }}\left( {{\rm{from normal tables}}} \right)\\\\ = 0.9938\\\end{array}$

(b)

The probability of X less than 70 is,

$\begin{array}{c}\\P\left( {X < 70} \right) = P\left( {\frac{{X - \mu }}{\sigma } < \frac{{70 - 100}}{{10}}} \right)\\\\ = P\left( {Z \le - 3} \right)\\\\ = 0.0013{\rm{ }}\left( {{\rm{from normal tables}}} \right)\\\end{array}$

(c)

Compute $P\left( {X < 80{\rm{ or }}X > 110} \right)$ .

$\begin{array}{c}\\P\left( {X < 80{\rm{ or }}X > 110} \right) = P\left( {X < 80} \right) + P\left( {X > 110} \right)\\\\ = P\left( {X < 80} \right) + \left[ {1 - P\left( {X < 110} \right)} \right]\\\\ = P\left( {\frac{{X - \mu }}{\sigma } < \frac{{80 - 100}}{{10}}} \right) + \left[ {1 - P\left( {\frac{{X - \mu }}{\sigma } < \frac{{110 - 100}}{{10}}} \right)} \right]\\\\ = P\left( {Z < - 2} \right) + \left[ {1 - P\left( {Z < 1} \right)} \right]\\\\ = 0.0228 + \left[ {1 - 0.8413} \right]\\\\ = 0.1814\\\end{array}$

(d)

Compute the two values between which 80% of the data falls.

Lower value:

$\begin{array}{c}\\P\left( {X < x} \right) = 0.10\\\\P\left( {\frac{{X - \mu }}{\sigma } < \frac{{x - 100}}{{10}}} \right) = 0.10\\\\P\left( {Z < \frac{{x - 100}}{{10}}} \right) = 0.10\\\\\frac{{x - 100}}{{10}} = - 1.2815\\\\x = 87.1845\\\end{array}$

Upper value:

$\begin{array}{c}\\P\left( {X \ge {x_2}} \right) = 0.10\\\\P\left( {\frac{{X - \mu }}{\sigma } \ge \frac{{{x_2} - 100}}{{10}}} \right) = 0.10\\\\P\left( {Z \ge \frac{{{x_2} - 100}}{{10}}} \right) = 0.10\\\\1 - P\left( {Z < \frac{{{x_2} - 100}}{{10}}} \right) = 0.10\\\\P\left( {Z < \frac{{{x_2} - 100}}{{10}}} \right) = 0.90\\\\\frac{{x - 100}}{{10}} = 1.2815\\\\x = 112.8155\\\end{array}$

Ans: Part a

The probability of X greater than 75 is 0.9938.

Part b

The probability of X less than 70 is 0.0013.

Part c

The probability of X less than 80 or greater than 110 is 0.1814.

Part d

80% of values fall between 87.1845 and 112.8155.