00:01
We know that the probability density function for the exponential distribution is fx equals minus, sorry, equals lambda times e to the power of minus nmadax.
00:23
And the cumulative distribution function equals 1 minus e to the power of minus nmdiacs.
00:34
So in part a, we want to find the value of the parameter lambda and we know that 10 % of components have filled by 1000 overs.
00:49
This means the probability that x is less than 1000 equals 0 .1.
01:00
So which is to say f 1000 equals 0 .1.
01:09
So 1 minus e.
01:13
To the power of minus nmda times 1 ,000 equals 0 .1.
01:21
And hence, e to the power of minus 1 ,000 nmda equals 0 .9.
01:31
So, minus 1 ,000 nmda equals log .9.
01:40
And we know that the mean equals 1 over nmada, and hence it is 1 ,000 divided by minus log .9.
02:00
The number is approximately 9 ,491 .22.
02:09
And the standard deviation is also one over numbed...