How to find how usual or unusual a probability is? guys. Just doing some homework and I came I across a problem that I did not know how to do, Could someone explain, Please and thank you.
Problem:
a group of 13 people is randomly selected to discuss products of the Yummy Company. It is determined that the mean number (per group) who recognize the Yummy brand name is 10.2 and the standard deviation is 0.84. Would it be unusual to randomly select 13 people and find that fewer than 7 recognize the Yummy brand name?
So I have the probability and standard deviation, however, I don't know what equation to use for this kind of problem? Could someone lend their expertise, please? Thank you.
 A: If you wanted to prove to yourself that fewer than 7 people recognize the brand name, you could employ a 'one sample t-test'. 
First, lets set up two hypotheses; $H_o$ and $H_a$. 
We interpret the alternative hypothesis $H_a$ as what the 'researcher' believes, which in this case is you. 
So, as the researcher you believe that the means are different. Not one greater than the other, just different, or 'unusual'. 
Therefore, 
$$H_a: \mu_A\ne \mu_B $$
Next, we can formulate $H_o$. Here $H_o$, which is the null hypothesis, is just the opposite of the alternative hypothesis $H_a$. So, the opposite of 'not equal' is really just, 'equal'. 
Therefore, 
$$H_o: \mu_A=\mu_B $$
Before we get into the nitty-gritty, lets think about what showing that a mean of $7$ is significantly different from $10.3$ would mean. 
If we showed that $7$ was significantly different from $10.3$, that would mean what? - It would mean that everything less than $7$ is also significantly different. 
Now we can use a couple of formulas to determine if the mean of the sample, which we will take as $7$, is significantly different from the accepted mean of $10.3$. 
The two formulas of interest will be; 
$$t = t_{\alpha , n-1}$$ and, 
$$t.s.v. = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$$
Here '$t$' represents our 'critical value', while '$t.s.v.$' represents our 'test statistic value'. (Hopefully you're familiar with what each of the variables above mean, and you're able to substitute the values appropriately.)
Once you evaluate $t$ and $t.s.v.$, you want to check which one is greater - which leads us to the final step in answering the question - the decision rule.  
Here, our decision rule will be, "If $t.s.v.>t$, accept $H_o$." Or in other words, the accepted sample mean ($10.3$) is not different to the new sample mean ($7$). 
A: A general test is 'number of standard deviations from the mean' with about 3.5 or more starting to be considered unusual -  in this case you have
$(10.2 - 7) / 0.84=3.8$ 
so it seems unusual
A: I understand that you have already turned in this paper because this thread is almost a year old, however, for anyone else out there having issues, here is a simple answer:
The formal definition of unusual is a data value more than 2 standard deviations away from the mean in either the positive or negative direction. 

Therefore, this will be your range of usual:
(0.84*2) + 10.2 = 11.88 this is your highest value
10.2 - (0.84*2) = 8.2 this is your lowest value
Usual range: $8.2 \le x \le 11.88$

Since 7 is less than your lowest end, 8.2, then it is definitely unusual.
