Binomial distribution - Question with exercise

I have the following exercise which they give us a hint that the binomial distribution is likely to apply.

A company manufactures car components. The quality control scheme for a particular type of component consists of taking random samples of six components at regular intervals. The number of defective components are counted and recorded. Out of 100 such samples, in 52 cases no defectives were found, in 34 cases one defective, in 10 cases two defectives and in the remaining cases three defectives.

Are these results consistent with the view that the process is operating with an average of 10% defectives?

In my understanding I need to use the following formula: $$$$P(x)=\frac{n!}{(n-x)!x!}p^xq^{n-x}$$$$

In 52 cases, 0 defectives: $$$$P(52)=\frac{100!}{(100-52)!52!}(0.1)^{52}(0.9)^{100-52}$$$$

Should I put $$p$$= 10% and $$q$$=90%. Because is says with an average of 10%. Or should I calculate $$p$$ and $$q$$ differently?

• You are correct to use $p=.1.$ But what is $n?$ What purpose is your last equation supposed to serve? It might help if we knew what topic you are studying currently. From what you say, there is no way to know whether you are supposed to do a chi-squared goodness-of-fit test to get your answer. Or whether you are just supposed to speculate if results 52, 34, 10, and 4 look 'pretty reasonable' (which I guess they do). – BruceET Apr 2 '20 at 17:25
• The question is what I highlighted in bold fond. To check whether the results are consistent with the view that the process is operating with an average of 10% defectives. They provided a hint that a binomial distribution is likely to apply that's why I thought to use that formula. But then, should I calculate the P for x=52, 34, 10 and 4? – Gina Apr 2 '20 at 17:41
• Sounds like you need to do a formal test. Getting probabilities for 52, 34, 10, and 4 is not the right approach. See my Answer. Also, maybe see examples in your text. (As I see it, my job is not to work the problem for you, but to nudge you in the right direction.) – BruceET Apr 2 '20 at 17:45
• Could you please explain how one can calculate Expected counts from the Binomial distribution? I thought was the expected value np but probably isn't right. – Gina Apr 2 '20 at 18:20

Then use the distribution $$\mathsf{Binom}(n=6, p=.1)$$ to find corresponding Expected counts (which need not be integers). My guess is that the Expected count for category '3 or more' may be below 5. In that case you'd need to use categories '0', '1', and '2 or more'.
If you do have three categories, the test statistic would be distributed approximately as $$\mathsf{Chisq}(\mathrm{df} = 3-1 = 2).$$ From that you can find the critical value of a test at level 5% or the P-value for your computed test statistic.
• I can't imagine a prohibition on using statistical software unless you're taking a remedial math class masquerading as a stat class. But I'm just using software for brevity in comments; 'dbinom(0, 6, .1)` is shorthand for $P(Y=0)$ where $Y \sim$ BINOM(n=6,p=.1). But with a little exploration, you might have figured that out. As I see it, my job is not to do your hwk for you, but to give you clues in what seems to me to be the right direction. If someone is going to do the problem for you step by step, that should be your instructor or a TA (whether in person or by distance learning). – BruceET Apr 2 '20 at 21:03