A random variable T has the following CDF: FT(t) = 1 -...

A random variable T has the following CDF: F_T(t) = 1 - e^{-10t^2} t ? 0 (a) What is P(T > 0.4)? (b) What is P(0.4 ? T ? 0.8)? (c) What is P(0.4 ? T ? 0.8 | T > 0.4)? (d) What is the probability density function for T? (e) What is the mean value of T? (f) What is the variance of T?

Transcript

00:01 Hello students in this question we have given a random variable t which has the cumulative distribution function that is cdf is 1 minus e raised to minus 10 into t square and t is greater than or equal to 0 then by using this in the first question we have to calculate probability of t is greater than 0 .4 so this is equal to 1 minus probability of t is less than or equal to 0 .4 and now this is 1 minus probability of t is less than or equal to 0 .4 is nothing but f of 0 .4 so now we have the cdf so this is equal to 1 minus 1 minus e raised to minus 10 into 0 .4 square so after simplifying this we will get the required probability is 0 .2019 then in the next part we have to calculate probability of 0 .4 is less than or equal to t is less than or equal to 0 .8 now this is equal to probability of t is less than or equal to 0 .8 minus probability of t is less than 0 .4 and now this is equal to f of 0 .8 minus f of 0 .4 and substituting these values in this cdf we will get 1 minus e raised to minus 10 into 0 .8 square minus 1 minus e raised to minus 10 into 0 .4 square and therefore the required probability is 0 .2002 then in the c part we have to calculate the conditional probability that is probability of 0 .4 is less than or equal to t is less than or equal to 0 .8 given that t is greater than 0 .4 now by using the formula for the conditional probability this probability is probability of 0 .4 is less than or equal to t is less than or equal to 0 .8 intersection t is greater than 0 .4 and divide by probability of t is greater than 0 .4 now the intersection of these two is probability of 0 .4 is less than or equal to t is less than or equal to 0 .8 at 0 .4 includes in this interval and this is divided by probability of t is greater than 0 .4 now we have already calculated these two probabilities which is 0 .2002 divided by 0 .2019 and therefore the required conditional probability is 0 .9916 and then in the d…

Question

A random variable T has the following CDF: F_T(t) = 1 - e^{-10t^2} t ? 0 (a) What is P(T > 0.4)? (b) What is P(0.4 ? T ? 0.8)? (c) What is P(0.4 ? T ? 0.8 | T > 0.4)? (d) What is the probability density function for T? (e) What is the mean value of T? (f) What is the variance of T?

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