Help on normal distribution question please

[gradds]

" A contractor has recently purchased a new bulldozer. On previous jobs, 15 out of a total of 50 bulldozers have broken down before the end of the job.

What is the mean and standard deviation of the probability distribution describing the probability of failure of a bulldozer?

Note: assume a normal approximation is applicable."

Mean of the probablity distribution seems obvious (0.3), but how does one find the standard deviation??

Thanks for any help that can be offered.

 

[PRodQuanta]

[tex]\sqrt{\frac{1}{n-1}\sum_{i=1}^n ({x_i}-{\bar{x}})^2}[/tex]
Hope this helps.

Paden Roder

 

[mathman]

In the standard dev. expression, the mean is 0.3, as you noted. The x_i are either 0 (35 terms) or 1 (15 terms), where n=50.

 

[HallsofIvy]

This is a binomial distribution (I'm not sure why they mention a "normal approximation"- if you were asked for specific probablities of, say, 17 breaking down, you might want to use an approximation but it isn't necessary just to find the mean and standard deviation).

In a binomial distribution with n occurances, probability p of "sucess" on anyone and (1-p) of "failure", the mean is np and the standard deviation is √(np(1-p)).

The probability of a single bulldozer breaking down, here, is p= 15/50= 0.3 and you are asked about 1 bulldozer so, yes, the mean is 0.3. The standard deviation is
√(0.3*0.7)= √(0.21).