Let 𝑋 be a continuous random variable with probability density function 𝑓 of 𝑥 equals four 𝑥 plus 𝑘 divided by 21 if 𝑥 lies between three and four inclusive and 𝑓 of 𝑥 equals zero otherwise. Find the value of 𝑘.
The integral of any probability density function is equal to one. This means that in our example, integrating four 𝑥 plus 𝑘 divided by 21 between the limits three and four will give us an answer of one. Integrating four 𝑥 gives us two 𝑥 squared and integrating 𝑘 gives us 𝑘𝑥. Therefore, the integral of four 𝑥 plus 𝑘 divided by 21 is two 𝑥 squared plus 𝑘𝑥 divided by 21.
We now need to substitute in the limits four and three and subtract the answers. Substituting in four gives us 32 plus four 𝑘 divided by 21 and substituting in three gives us 18 plus three 𝑘 divided by 21. Multiplying each term in this equation by 21 gives us 32 plus four 𝑘 minus 18 plus three 𝑘 equals 21.
Grouping the like terms or simplifying the left-hand side of the equation gives us 14 plus 𝑘 as 32 minus 18 is 14 and four 𝑘 minus three 𝑘 is equal to 𝑘. Finally, if we subtract 14 from both sides of this equation, we’re left with 𝑘 is equal to seven as 21 minus 14 is equal to seven. This means that when the probability density function of a continuous random variable is 𝑓 of 𝑥 is equal to four 𝑥 plus 𝑘 divided by 21 between three and four, then 𝑘 is equal to seven.
Therefore, 𝑓 of 𝑥 is equal to four 𝑥 plus seven divided by 21 if 𝑥 lies between three and four and 𝑓 of 𝑥 equals zero otherwise.