Binomial Distribution for Manufacturer's Claim on Product Durability

[dfraser]

Hi,

I'm struggling to know what distribution this question requires, and what should be signalling the distribution type:

A manufacturer claims at most 5% of his product will sustain fewer than 1000hrs of operation before needing service. Twenty products are selected at random from the production line and tested. It was found that 3 of them required service before 1000hrs of operation. Comment on the manufacturers claim.

The solution for the probability of 3 requiring service before the 1000hrs is (according to my text) 0.0754.

I think I might be getting confused by "at most 5% of his product will sustain fewer than 1000hrs".

For a binomial I'd have:
$${ 20\choose17 }{.95}^{17}{.05}^{3} = .05958$$

For a negative binomial I'd have:
$${ 19\choose2 }{.95}^{17}{.05}^{3} = .00893$$

This isn't poisson, geometric, hypergeometric, or multinomial, so what is it? Or can someone give me a hint about where I'm going wrong?

 

[I like Serena]

Hi,

I'm struggling to know what distribution this question requires, and what should be signalling the distribution type:

A manufacturer claims at most 5% of his product will sustain fewer than 1000hrs of operation before needing service. Twenty products are selected at random from the production line and tested. It was found that 3 of them required service before 1000hrs of operation. Comment on the manufacturers claim.

The solution for the probability of 3 requiring service before the 1000hrs is (according to my text) 0.0754.

I think I might be getting confused by "at most 5% of his product will sustain fewer than 1000hrs".

For a binomial I'd have:
$${ 20\choose17 }{.95}^{17}{.05}^{3} = .05958$$

For a negative binomial I'd have:
$${ 19\choose2 }{.95}^{17}{.05}^{3} = .00893$$

This isn't poisson, geometric, hypergeometric, or multinomial, so what is it? Or can someone give me a hint about where I'm going wrong?

Hi dfraser,

It's a binomial distribution as you surmised.
It's just that they ask for $P(X\ge 3)$ instead of $P(X=3)$, where $X$ is the number that requires service before 1000 hrs.
And indeed we assume that the probability is 5% per product, since this gives the manufacturer the most leeway.