# Help with Normal Distribution question!Watch

#1
Hi...I've been trying to do this question for ages and its not happening!

The first part is:
a) It is known that 60% of families in the UK have video recorders. If one takes a sample of 5 families what is the chance that at least 80% of them (i.e. 4 or 5 families) have a video recorder?

I've done this part and got 0.33696 using the binomial distribution.
n=5, p=0.6, r=0.4 and nCr pr qn-r

so P(4 in 5 successes) + P(5 in 5 successses) = 0.2592 + 0.07776 = 0.33696

Second part (part I can't do):
b) If the sample in part (a) was 1000 familes, why might you assume the number of families in the sample with video recorders is normally distributed? What is the mean and the variance of this normal distribution. What is the chance more than 800 (more than 80%) have video recorders?

Many Thanks

Emma
0
13 years ago
#2
For large n, the normal distribution is a good estimate of the binomial distribution.

n=1000, p=0.6, mean=1000x0.6=600, var=1000x0.6x0.4=240
X~N(600,240)

P(X>800) = (standardising) P(Z>(800-600)Ã·240) where Z~N(0,1)

P(X>800) = P(Z>(5/6)) = 1 - P(Z<=(5/6)) = 1 - 0.7967 (from tables) = 0.2033

0
#3
(Original post by Applying4Maths)
For large n, the normal distribution is a good estimate of the binomial distribution.

n=1000, p=0.6, mean=1000x0.6=600, var=1000x0.6x0.4=240
X~N(600,240)

P(X>800) = (standardising) P(Z>(800-600)Ã·240) where Z~N(0,1)

P(X>800) = P(Z>(5/6)) = 1 - P(Z<=(5/6)) = 1 - 0.7967 (from tables) = 0.2033

So what value do I look up in the tables? where does the 0.7967 come from? I looked up 5/6 to get 0.2033....where does the 0.7967 come from?? Many thanks though, your a lifesaver!
0
13 years ago
#4
On the tables for ?(z) in the formulae book, 0.83 (5/6) --> 0.7967

as this is for Z<=(5/6), you need to do 1 - 0.7967 to find the value you are looking for, 0.2033

Maybe if you are doing a different exam board you have different tables?
In my formulae book (EdExcel) you only get values for P(Z<=z), no values for P(Z>z)

Laura
0
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