A farmer finds that the weights of sheep on his farm have a normal distribution with mean 66.4 kg
and standard deviation 5.6 kg.
250 sheep are chosen at random. Estimate the number of sheep which have a weight of between
70 kg and 72.5 kg.
Why there is no use of continuity correction in this question?And please show full solution of the question.
This question is referring to your z-scores : )
You need your z score formula z=(x- mean/standard deviation
You will find your z score for 72.5 and then a second z-score of 70
and since the z-score table reads "less than" you will need to subtract the value of z found at 70 from the value of z found at 72.5
This will give you the percent between 70 adn 72.5
then you will have to multiply that percent by your 250 sheep to see how many would be between those weights.
Your job is to follow these steps and produce a solution.
Why can't we use P((72-66.4)/5.6) -(P(69.5-66.4)/5.6)?
If you are still using "tables", take a look at this webpage,
which is probably the best for your type of question.
http://davidmlane.com/normal.html
You can bypass the z-score calculations and use the data
directly as it is given to you, although the z-scores you calculated
can be used as well. As a matter of fact , the default is a mean of 0 and
a SD of 1
Using the data as is, the result is .1221
I clicked on the 'between' button then entered 70 and 72.5
Using the z-scores that "Ms Pi" outlined, you would have....
for 70: z-score = (70-66.4)/5.6 = .642857...
and from your table you should get P(z < 70) = 0.7398
for 72.5" z-score = (72.5-66.4)/5.6 = 1.08928..
and from your table you should get P(z < 72.5) = .862
so P( 70 < x < 72.5) = .862 - .7398 = .1222, same as before
You then ask: Why can't we use P((72-66.4)/5.6) - (P(69.5-66.4)/5.6)??
I don't know where you are getting your last P(69.5-66.4)/5.6) from?
Is 69.5 a typo?
Anyway, that is in effect what I did, let me know if this made sense to you.
Why do think it is a typo?I made that step intentionally.
And for the P((72-66.4)/5.6), I am getting probability 1.
makes no sense...
for 72.5" z-score = (72.5-66.4)/5.6 = 1.08928..
Now find 1.08928 in your table (or use my webpage)
Here is a table:
https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf
and from your table you should get P(z < 72.5) = .862
The table I gave you should give you .86214, but I had to round 1.08928 to 1.09
As normal approximation to binomial,shouldn't it be P(X<72) for P(X<72.5) and P(X<69.5) for P(X<70) while continuity correction?
"As normal approximation to binomial,shouldn't it be P(X<72) for P(X<72.5) and P(X<69.5) for P(X<70) while continuity correction?"
Must admit that I don't understand what that procedure is.
The last time I taught this, 35 years ago, I never came across that terminology.
Did your text give you an answer to this question? Was my answer not correct?
I just noticed that your same question was properly answered for you at
https://www.jiskha.com/questions/1820564/a-farmer-finds-that-the-weights-of-sheep-on-his-farm-have-a-normal-distribution-with-mean
by two capable tutors.
As was pointed out at the end of these replies, there is no need to mess around with
interpolation between values not given by tables.
Tables are a thing of the past and the webpage that both oobleck and I gave you
is extremely precise, and far superior to any tables.
Btw, my answer is correct.