Question

If bolt thread length is normally distributed, what is the probability that the thread length of a randomly selected bolt is

a) Within 1.5 SDs of its mean value

b) Farther than 2.5 SDs from its mean value

c) Between 1 and 2 SDs from its mean value

Answer 1

Concepts and reason

The concept of normal distribution and standard normal variates are used to calculate the probability.

The normal distribution can be used to find the probability of the continuous variable when the data is more or less symmetric. The probability of being less than or more than some value can be calculated by the area of the curve to the left of that value.

Fundamentals

A continuous random variable is said to follow a normal distribution if the probability density function can be written in the form:

$f\left( x \right) = \left\{ \begin{array}{l}\\\frac{1}{{\sigma \sqrt {2\pi } }}{e^{ - {{\left( {\frac{{x - \mu }}{\sigma }} \right)}^2}}}{\rm{ }} - \infty \le x \le \infty \\\\0{\rm{ Otherwise}}\\\end{array} \right.$

Here, $\mu$ is the mean and $\sigma$ is the standard deviation of the distribution. The probability of normal distribution can be defined as:

$P\left( {x > \alpha } \right) = \int\limits_\alpha ^\infty {\frac{1}{{\sigma \sqrt {2\pi } }}} {e^{ - {{\left( {\frac{{x - \mu }}{\sigma }} \right)}^2}}}dx$

To calculate the probability of a random variable $N\left( {\mu ,{\sigma ^2}} \right)$ , convert the variable to a standard normal variable by the transformation,

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

$\mu$ is the mean and $\sigma$ is the standard deviation of the distribution.

Calculate the probability using the formula:

$\begin{array}{c}\\P\left( {a < x < b} \right) = P\left( {\frac{{a - \mu }}{\sigma } \le \frac{{x - \mu }}{\sigma } \le \frac{{b - \mu }}{\sigma }} \right)\\\\ = P\left( {\frac{{a - \mu }}{\sigma } \le z \le \frac{{b - \mu }}{\sigma }} \right)\\\\ = \Phi \left( {\frac{{b - \mu }}{\sigma }} \right) - \Phi \left( {\frac{{a - \mu }}{\sigma }} \right)\\\end{array}$

To calculate the probability where the interval is $- a < x < a$ , use the formula:

$\begin{array}{c}\\P\left( { - a < x < a} \right) = P\left( {\frac{{ - a - \mu }}{\sigma } \le \frac{{x - \mu }}{\sigma } \le \frac{{a - \mu }}{\sigma }} \right)\\\\ = P\left( {\frac{{ - a - \mu }}{\sigma } \le z \le \frac{{a - \mu }}{\sigma }} \right)\\\\ = 2\Phi \left( {\frac{{a - \mu }}{\sigma }} \right) - 1\\\end{array}$

Use the formula $= {\rm{NORMSDIST}}\left( {} \right)$ to calculate the probability for the Z-value.

(a)

A bolt thread’s length $\left( x \right)$ is normally distributed. The thread length of a randomly selected bolt lies $1.5\sigma$ of its mean intercept, as the thread length lies between $\mu - 1.5\sigma$ and $\mu + 1.5\sigma$ . The probability is calculated as:

$\begin{array}{c}\\P\left( {\mu - 1.5\sigma \le x \le \mu + 1.5\sigma } \right) = P\left( {\frac{{\left( {\mu - 1.5\sigma } \right) - \mu }}{\sigma } \le \frac{{x - \mu }}{\sigma } \le \frac{{\left( {\mu + 1.5\sigma } \right) - \mu }}{\sigma }} \right)\\\\ = P\left( { - 1.5 \le z \le 1.5} \right)\\\\ = 2\Phi \left( {1.5} \right) - 1\\\end{array}$

Use Excel to calculate the probability for the Z-Value 1.5. The screenshot of the formula used is shown:

Thus,

$\begin{array}{c}\\P\left( {\mu - 1.5\sigma \le x \le \mu + 1.5\sigma } \right) = 2 \times 0.9331928 - 1\\\\ = 0.8663856\\\\ \approx 0.8664\\\end{array}$

(b)

A bolt thread’s length $\left( x \right)$ is normally distributed. The thread length of a randomly selected bolt is lain further than $2.5\sigma$ of its mean.

The probability is calculated as:

$\begin{array}{c}\\P\left( {x < \mu - 2.5\sigma {\rm{ or }}x > \mu + 2.5\sigma } \right) = 1 - P\left( {\mu - 2.5\sigma \le x \le \mu + 2.5\sigma } \right)\\\\ = 1 - P\left( {\frac{{\mu - 2.5\sigma - \mu }}{\sigma } \le \frac{{x - \mu }}{\sigma } \le \frac{{\mu + 2.5\sigma - \mu }}{\sigma }} \right)\\\\ = 1 - P\left( { - 2.5 \le z \le 2.5} \right)\\\\ = 1 - \left( {2\Phi \left( {2.5} \right) - 1} \right)\\\end{array}$

Use Excel to calculate the probability for the Z-Value 2.5. The screenshot of the formula used is shown:

So,

$\begin{array}{c}\\P\left( {x < \mu - 2.5\sigma {\rm{ or }}x > \mu + 2.5\sigma } \right) = 1 - \left( {\left( {2 \times 0.9937903} \right) - 1} \right)\\\\ = 1 - 0.9875806\\\\ = 0.0124194\\\\ \approx 0.0124\\\end{array}$

(c)

The calculation of the probability of the thread length is between the $\sigma$ and $2\sigma$ of its mean:

$\begin{array}{c}\\P\left( {\mu - 2\sigma \le x \le \mu + 2\sigma {\rm{ or }}\mu - \sigma \le x \le \mu + \sigma } \right) = \left( \begin{array}{c}\\P\left( {\mu - 2\sigma \le x \le \mu + 2\sigma } \right) - \\\\P\left( {\mu - \sigma \le x \le \mu + \sigma } \right)\\\end{array} \right)\\\\ = \left( {2\Phi \left( 2 \right) - 1} \right) - \left( {2\Phi (1) - 1} \right)\\\end{array}$

Use Excel to calculate the probability for the Z-Value 1 and 2. The screenshot of the formula used is shown below:

And:

Thus,

$\begin{array}{c}\\P\left( {\mu - 2\sigma \le x \le \mu + 2\sigma {\rm{ or }}\mu - \sigma \le x \le \mu + \sigma } \right) = \left( {\left( {2 \times 0.97725} \right) - 1} \right) - \left( {\left( {2 \times 0.841345} \right) - 1} \right)\\\\ = \left( {1.9545 - 1} \right) - \left( {1.682689 - 1} \right)\\\\ = 0.9545 - 0.682689\\\\ = 0.271811\\\end{array}$

Ans: Part a

The probability that the thread length of a bolt lies within $1.5\sigma$ of its mean is approximately 0.8664.

Part b

The probability that the thread length of a bolt is further than $2.5\sigma$ of its mean is approximately 0.0124.

Part c

The probability of the bolt thread length between $\sigma$ and $2\sigma$ from its mean value is 0.271811.