Question

One piece of PVC pipe is to be inserted inside another piece. The length of the first piece is normally distributed with mean value \({\rm{20}}\) in. and standard deviation \({\rm{.5}}\) in. The length of the second piece is a normal \({\rm{rv}}\)with mean and standard deviation \({\rm{15}}\) in. and \({\rm{.4}}\) in., respectively. The amount of overlap is normally distributed with mean value \({\rm{1}}\) in. and standard deviation \({\rm{.1}}\) in. Assuming that the lengths and amount of overlap are independent of one another, what is the probability that the total length after insertion is between \({\rm{34}}{\rm{.5 }}\) in. and \({\rm{35}}\) in.?

Short answer

\({\rm{P}}\left( {{\rm{34}}{\rm{.5\pounds}}{{\rm{T}}_{\rm{0}}}{\rm{\pounds35}}} \right){\rm{ = 0}}{\rm{.1588}}\)

Step-by-step solution

Definition of standard deviation

The square root of the variance is the standard deviation of a random variable, sample, statistical population, data collection, or probability distribution. It is less resilient in practice than the average absolute deviation, but it is algebraically easier.

Calculating the probability that the total length after insertion is between \({\rm{34}}{\rm{.5 in}}{\rm{. and 35 in}}{\rm{.}}\)

Use the random variables \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}\) to represent the three component lengths. The entire length is going to be

\({{\rm{T}}_{\rm{0}}}{\rm{ = }}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}{\rm{ - }}{{\rm{X}}_{\rm{3}}}{\rm{,}}\)

since the overlapping half is represented by the third random variable, \({{\rm{X}}_{\rm{3}}}\)

Standardize the random variable in order to calculate the requested probability. The average number is

\(\begin{aligned} {{{\rm{\mu }}_{{{\rm{T}}_{\rm{0}}}}} &= E \left( {{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}{\rm{ - }}{{\rm{X}}_{\rm{3}}}} \right) &= E\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ + E}}\left( {{{\rm{X}}_{\rm{2}}}} \right){\rm{ - E}}\left( {{{\rm{X}}_{\rm{3}}}} \right)}\\ &= {\rm{20 + 15 - 1 = 34}\end{aligned}\)

the variance is

\(\begin{aligned} {{\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}^{\rm{2}} &= V\left( {{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}{\rm{ - }}{{\rm{X}}_{\rm{3}}}} \right)\ &= {\rm{V}}\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ + V}}\left( {{{\rm{X}}_{\rm{2}}}} \right){\rm{ + ( - 1}}{{\rm{)}}^{\rm{2}}}{\rm{V}}\left( {{{\rm{X}}_{\rm{3}}}} \right)}\\&= {\rm{0}}{\rm{.25 + 0}}{\rm{.16 + 0}}{\rm{.01}}}\\ & = 0 {\rm{.42,}}}\end{aligned}\)

(1) In the exercise, mean values and standard deviations are provided.

(2): an unrelated random variable

The standard deviation of the data is

\({{\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}}{\rm{ = }}\sqrt {{\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}^{\rm{2}}} {\rm{ = 0}}{\rm{.6481}}\)

The likelihood of the event \(\left\{ {{\rm{34}}{\rm{.5£ }}{{\rm{T}}_{\rm{0}}}{\rm{£ 35}}} \right\}\)is

\(\begin{array}{*{20}{c}}{{\rm{P}}\left( {{\rm{34}}{\rm{.5£}}{{\rm{T}}_{\rm{0}}}{\rm{£35}}} \right){\rm{ = P}}\left( {\frac{{{\rm{34}}{\rm{.5 - 34}}}}{{{\rm{0}}{\rm{.6481}}}}{\rm{£}}\frac{{{{\rm{T}}_{\rm{0}}}{\rm{ - }}{{\rm{\mu }}_{{{\rm{T}}_{\rm{0}}}}}}}{{{{\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}}}}{\rm{£}}\frac{{{\rm{35 - 34}}}}{{{\rm{0}}{\rm{.6481}}}}} \right)}\\{{\rm{ = P(0}}{\rm{.77£ Z£ 1}}{\rm{.54) = P(Z£1}}{\rm{.54) - P(Z£0}}{\rm{.77)}}}\\{\mathop {\rm{ = }}\limits^{{\rm{(3)}}} {\rm{0}}{\rm{.1588}}}\end{array}\)

(3): from the appendix's normal probability table. Software can also be used to calculate the likelihood.