Machine Life random variable with probability...

$\begin{array}{l}{\text { Machine Life }} \\ {\text { random variable with probability density function defined }} \\ {\text { by }}\end{array}$ $$f(t)=\frac{1}{11}\left(1+\frac{3}{\sqrt{t}}\right) \text { for } t \text { in }[4,9].$$ $\begin{array}{l}{\text { (a) Find the mean life of this machine. }} \\ {\text { (b) Find the standard deviation of the distribution. }} \\ {\text { (c) Find the probability that a particular machine of this }} \\ {\text { kind will last longer than the mean number of }} \\ {\text { years. }}\end{array}$

$\begin{array}{l}{\text { Machine Life }} \\ {\text { random variable with probability density function defined }} \\ {\text { by }}\end{array}$
$$f(t)=\frac{1}{11}\left(1+\frac{3}{\sqrt{t}}\right) \text { for } t \text { in }[4,9].$$
$\begin{array}{l}{\text { (a) Find the mean life of this machine. }} \\ {\text { (b) Find the standard deviation of the distribution. }} \\ {\text { (c) Find the probability that a particular machine of this }} \\ {\text { kind will last longer than the mean number of }} \\ {\text { years. }}\end{array}$

Key Concepts

Variance and Standard Deviation

Variance measures the spread of a probability distribution by computing the expected squared deviation from the mean. The standard deviation, which is the square root of the variance, provides a scale for the dispersion of the values around the mean, offering insight into the reliability and variability of the outcomes.

Integration Techniques for Continuous Distributions

Integration is a fundamental technique for working with continuous distributions, enabling the calculation of probabilities, expected values, and variances. Mastery of integration techniques, including substitution and handling functions involving radicals or other transformations, is essential for accurately evaluating these quantities over the specified intervals.

Probability Density Function

A probability density function (PDF) describes how the probability of a continuous random variable is distributed over its range. It is a nonnegative function that integrates to one over its support, and probabilities for events are determined by integrating the PDF over the desired interval.

Expected Value

The expected value, or mean, of a continuous random variable is the weighted average of all possible values, calculated by integrating the product of the variable and its probability density function over its range. This measure provides a central tendency that summarizes the distribution.

Transcript

00:04 Number 25 our probability density function is f of t and that even integral is core to 9 now first we need to find main value mu so mean value is given by mu is equal to integration of a 2b t into f of d d d now put the value of f of t which is 1 by 11 1 plus 3 under root t here 1 by 11 is constant so we can write outside of the integral here t into 1 is t plus 3 here t -d s to 1 and in the denominator we have t -d s to 1 by 2 so 1 minus 1 by 2 is 1 by 2 now take integration of t which is t squared t by 2 2 plus 3 and integration of t -d s 2 1 by 2 is 3 3 by 2 upon 3 by 2 here 3 and 3 will be cancelled out so right now execute upper limit 9 minus lower limit is 4 so here upper limit is 9 squared divided by 2 plus 2 9 dash 2 3 by 2 and low limit is 4 so here 4 squared divided by 2 minus 2 4 to 3 by 2 so 9 square is 81 divided by 2 plus 54 minus 8 minus 16 so here mu is 141 divided by 22 which is approximate to 6 .40 409 which is our expected value or we can say mean value now second thing is we need to find variants which is given by integration of a 2b x square or we can say t squared so we have variable t so we need to write t so here t square f of t d d minus mu square again with the value of f of t which is 1 by 11 plus 3 upon under t and mu is 6 .409 square which we have already found in part 1 again here 1 by 11 is constant so t squared into 1 is t square so here we have t rest to 2 and in denometer we have t x 2 1 by 2 so 2 minus 1 by 2 is 3 by 2 integration of t squared is tq divided by 3 and integration of t dash to 3 by 2 is t to 5 by 2 upon 5 by 2 now 2 into 3 is 6 and in the perimeter we have 5 so you have 6 by 5 t is to 5 by 2 upon 1 0 0 square now execute upper limit minus flawed limit so variance is equal to 1 by 11 upper limit is 9 so here 9 rest to 3 upon 3 plus 6 by 5 9 rest to 5 by 4 sorry 5 by 2 minus lower limit is 4 so here 4 2 divided by 3 minus 6 by 5 5 4 rest to 5 by 2 minus 6 0 0 0 9 square as it is so here 3 is equal to 2 .094 which is our variance now standard deviation sigma is given by square root of variance v so here square root of 2 .094 and we get our standard deviation sigma is 1 .447 okay now third step is we need to find the probability that a particular machine of this kind will last longer than the mean number of years so here mean is 6 .409 less than equal to 2 and less than equal to our upper limit is 9 so for integration our limits of 6 .409 to 9 so probability integration of 6 .409 to 9 so probability integration of 6 .409 to 9 f of x x is 1 by 11 1 plus 3 upon under t t again here 1 by 11 is constant 1 plus 3 here t under root t which is denominator so here you can write changes to minus 1 by 2 d t now integration of 1 is simple t and integration of t -res 2 minus 1 by 2 is t -res 2 1 by 2 of all 1 by 2.

07:19 Here 2 into 3 is 6...

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