Cereal boxes (normal distribution)

binomial distributionnormal distributionprobabilitystatistics

In the first part of the assignment, there was a surprise gift in every 7 cereal boxes. The company ships 300 cereal boxes to each store. I had to calculate the lower and upper bound (where is the limit to suspect non-random distribution of gifts).

I have trouble with the second part of the assignment.

  1. We assume that the surprise gifts in cereals are randomly distributed. How many cereals do you have to buy, with a probability of 0.8, to get at least 50 surprise gifts?

a) Calculate the number using normal distributions.

I took the n=300 from the first part.

p=0.8

  • Expected value= n*p=240

  • Standard deviation(sd)= sqrt(np(1-p))= 6.9

P(X=> 50)

At first, I calculated:

z-score = (50-240)/6.9 = -27.42414

Because of the high number of the z score, I decided to ignore the number and try to calculate the number this way: I looked at the z table (the probability is .8000, so the closest value in z table is 0.84).

$z= \frac{(x – \mu)}{sd}$

Solve x:

x = z*sd+ex

x=245.8

Then for some reason, I decided to subtract from n:
n-x = 54

I know this doesn't make any sense…

There is also the third part, that I cannot solve because of the unfinished second part.

Get the number (using R):

  • By simulating 10.000 random samples from true distributions,

  • Using quantiles/binomial probability

What is meant by the true distribution? I understand that as a normal distribution…

Thank you for your time and help.

Best Answer

The probability of 0.8 help you get the $z$ score $z=0.84$.

Then the probability of getting a toy is still $p=\frac17$. We are now looking for the value of $n$ which will give us $$z=\frac{x-\mu}{\text{sd}}$$ $$0.84=\frac{50-n\frac17}{\sqrt{n\frac17\frac67}}$$ We just need to solve for $n$. Can you take it from here?

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