Calculate Probability

[BrandonP]

Hello I am new to this website and wish to get some help on these questions regarding Binomial Distribution. See Below;

Specifically I need help with question (a&b.) I just can't seem to get my head round how these questions work and to how to even check my answer is correct.
Please see my workings out so far for question (b), I have no idea if I am on the right path, but any advice would be great.

 

[Steven G]

Part a) I guess you do not understand what % means.

% means out of 100.

So 85.4% of the washers meet tolerance levels means that 85.4 out of 100 meet the tolerance level. So how many do you thing will meet the tolerance level out of 50?


Part b) What you have to the left of all the equals signs are correct. I can only assume the results on the right hand are correct. You added them all up and the sum you got seems correct.

So what does this sum represent? Is that what you are trying to find?

Please post back?

 

[Dr.Peterson]

Hello I am new to this website and wish to get some help on these questions regarding Binomial Distribution. See Below;
View attachment 19250
Specifically I need help with question (a&b.) I just can't seem to get my head round how these questions work and to how to even check my answer is correct.
Please see my workings out so far for question (b), I have no idea if I am on the right path, but any advice would be great.
View attachment 19251

Is there a reason you changed 85.4% to 85% in all your work? You should be using the data you are given, and never round until the end.

Pretending it was really 85%, your P(0) is still wrong. It's useful to realize that in the formula you are using, the sum of the exponents is always n.

But there's a bigger issue. The way you chose p, it is the probability of a defective washer, so you've found the probability that 5 or fewer are defective. That's not what you were asked for! Always define your events and variables.

You didn't show any thinking for (a), so I'm not sure how you're doing there.

 

[BrandonP]

Hello, sorry for the late response have only saw both of your replies. These are probably easy questions but I really struggle with maths, but I will try my best!
Part a) I assume P=0.854, then to find Q=1-0.854=0.146, then I guess to find how many faulty washers = 0.146x50=7.3? (But you can't have 7.3 washers, so do you just round down?) (N=number of samples)
Part b) I was thinking to work out more than 5, you would first find out the value of (x<5), then to remove the variables you don't need, (x>5) = 1-(x<5), am I right in saying this is the correct way to go but obviously use the full values of p and q not rounding them.

 

[BrandonP]

Is there a reason you changed 85.4% to 85% in all your work? You should be using the data you are given, and never round until the end.

Pretending it was really 85%, your P(0) is still wrong. It's useful to realize that in the formula you are using, the sum of the exponents is always n.

But there's a bigger issue. The way you chose p, it is the probability of a defective washer, so you've found the probability that 5 or fewer are defective. That's not what you were asked for! Always define your events and variables.

You didn't show any thinking for (a), so I'm not sure how you're doing there.

Please read response above thanks.

 

[Dr.Peterson]

Hello, sorry for the late response have only saw both of your replies. These are probably easy questions but I really struggle with maths, but I will try my best!
Part a) I assume P=0.854, then to find Q=1-0.854=0.146, then I guess to find how many faulty washers = 0.146x50=7.3? (But you can't have 7.3 washers, so do you just round down?) (N=number of samples)
Part b) I was thinking to work out more than 5, you would first find out the value of (x<5), then to remove the variables you don't need, (x>5) = 1-(x<5), am I right in saying this is the correct way to go but obviously use the full values of p and q not rounding them.

(a) Your answer is correct. Mathematically, an expected value need not be a possible value, in much the same way we can say that the average family has 2.4 children, or whatever. It's an average, and does not reflect what you expect on any one batch, much less on every batch of 50. But practically speaking, you could say "I expect about 7 or 8 to be faulty in the batch." But your teacher (or an example in your textbook) can tell you what sort of answer is expected. I expect that 7.3 would be the expected answer.

(b) Yes, you're right, and I was guessing that you just hadn't shown your final answer. Of course, what you mean (but couldn't type easily here) was that in order to find P(x>5), you found 1 - P(x≤5). You could type that as 1 - P(x<=5) to avoid having to paste the symbol from somewhere else.

 

[BrandonP]

(a) Your answer is correct. Mathematically, an expected value need not be a possible value, in much the same way we can say that the average family has 2.4 children, or whatever. It's an average, and does not reflect what you expect on any one batch, much less on every batch of 50. But practically speaking, you could say "I expect about 7 or 8 to be faulty in the batch." But your teacher (or an example in your textbook) can tell you what sort of answer is expected. I expect that 7.3 would be the expected answer.

(b) Yes, you're right, and I was guessing that you just hadn't shown your final answer. Of course, what you mean (but couldn't type easily here) was that in order to find P(x>5), you found 1 - P(x≤5). You could type that as 1 - P(x<=5) to avoid having to paste the symbol from somewhere else.

Thank You very much for your helpful advice Dr.Peterson

I managed to get question (a) correct ages ago, but really needed a second opinion on the value I got.
As for question (b), my main problem was the fact I rounded off the p and q values and I now realize thats was caused me to go wrong in the first place. Once I used the correct values, I found that P(x≤5)= 0.2426, then I did 1 - P(x≤5)=0.7574 (75.74%)

You pointing out my problem with the P value made me realize the error I had made.