normal distribution question?

Question

Hi,
I was doing a question on normal distribution:

The lifetime of a transistor: with mean 1200 and standard deviation 100 (hours). I calculated the probability that the lifetime was greater than 1250 hrs was 0.3085

A question was then asked that if there are now 5 new transistors which are turned on at the same time, what is the probability that at least 4 of them are still working after 1250 hours.

How would you do this? (and also, if you could do this to show me how its done would be great)

Thanks.

Answer 1

First your work with the normal is correct. well done.



Let X be the number of working transistors after 4 hours. X has the binomial distribution with n = 5 trials and success probability p = 0.3085



In general, if X has the binomial distribution with n trials and a success probability of p then

P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)

for values of x = 0, 1, 2, ..., n

P[X = x] = 0 for any other value of x.



The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.

Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.



X ~ Binomial( n , p )



the mean of the binomial distribution is n * p = 1.5425

the variance of the binomial distribution is n * p * (1 - p) = 1.066639

the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.032782



The Probability Mass Function, PMF,

f(X) = P(X = x) is:



P( X = 0 ) = 0.1581106

P( X = 1 ) = 0.3526906

P( X = 2 ) = 0.3146929

P( X = 3 ) = 0.1403944

P( X = 4 ) = 0.03131720

P( X = 5 ) = 0.002794318





The Cumulative Distribution Function, CDF,

F(X) = P(X ≤ x) is:



x

∑ P(X = t) =

t = 0



P( X ≤ 0 ) = 0.1581106

P( X ≤ 1 ) = 0.5108012

P( X ≤ 2 ) = 0.825494

P( X ≤ 3 ) = 0.9658885

P( X ≤ 4 ) = 0.9972057

P( X ≤ 5 ) = 1





1 - F(X) is:



n

∑ P(X = t) =

t = x



P( X ≥ 0 ) = 1

P( X ≥ 1 ) = 0.8418894

P( X ≥ 2 ) = 0.4891988

P( X ≥ 3 ) = 0.1745059

P( X ≥ 4 ) = 0.03411151 = P(X = 4) + P(X = 5) ← answer

P( X ≥ 5 ) = 0.002794318

Answer 2

The Binomial distribution has the probability

function P(x=r)= nCr p^r (1-p)^(n-r)

r=0,1,2,.....,n

where nCr = n! / r! (n-r)!

You'll have to use the Binomial distribution in this case with n=5, p=0.3085 and r=4,r=5 and add them.

P(r=4)=5C4 (0.3085)^4(1-0.3085)^1

P(r=5)=5C5(0.3085)^5

=0.0313+0.0028=0.0341

You must have access to a Binomial table. You may not find p=0.3085. You have two options. Either compute it using a calculator or use a p=0.30 (close enough to 0.3085)

Answer 3

I only discovered this notation about this.... "the traditional distributions are a serious classification of statistical distributions. All common distributions are symmetric and function bell-formed density curves with a unmarried top".

Answer 4

.3085^4



like rolling a dice the probability of rolling a 6 is 1/6 and to six's is 1/6 * 1/6 is 1/36



same concept