Probability of an event occurring more than X% of the time

Question

Cans are a normal random variable with a mean of 7.96 ounces and a standard deviation of 0.22 ounces. Suppose that you draw a random sample of 34 cans.

Use normal approximation to find the probability that more than 48% of the sampled cans exceeds 8 ounces.

The first thing I did was find the probability that a can is over 8 ounces, which I got as 0.145.

The next thing I tried was to do was have it has a binomial distribution since its either overweight or not, but that was wrong. What should I do to get the correct probability?

Thank you for any help understanding how to approach this.

Answer 1

'The general procedure looks correct. After calculating the probability that a can is over $8$ ounces, you might want to use the normal to approximate the binomial.

Your "$p$" for the binomial is not right. We want $\Pr\left(Z\gt \frac{0.04}{0.22}\right)$. This is much larger than $0.145$.

Answer 2

I'm just going to answer the question from the start because I imagine that the original asker has moved on. However, someone has likely stumbled across this looking to solve the problem.

The first step is to solve for the probability that a can exceeds 8 ounces.

This can be done using the normalcdf function (which can be found under 2nd distr on a TI-84)

the lower limit is: 8

the upper limit is: inf. or 1E99

the mean is: 7.96

the standard deviation is: 0.22

In the calculator, this would look like: normalcdf(8,1E99,7.96,0.22) = .428

Once again, this is just the probability that a can will be between 8 and inf. ounces given that the mean is 7.96 and the standard dev. is 0.22.

Following this, you want to use the normalcdf to approximate the binomial distribution which is mentioned in the other comment.

This is once again done using the normalcdf function.

the lower limit is: 34 * 0.48 ................................(n * .48)

the upper limit is: 34 .......................................(n)

the mean is: 34 * 0.428 ......................................(n * p)

the standard deviation is: sqrt(34 * 0.428 * (1 - 0.428)) ....(sqrt(n * p * q), q=(1-p))

// explanations

Basically, the lower limit is the minimum amount of cans that must be greater than 8 ounces (34 * .48). This value will not be an integer because we are approximating it with a continuous distribution (normalcdf). The upper limit is 34 because the sample size is 34, so we can't have 35 cans that weigh over 8 ounces because we only have 34 cans. The mean is n * p which is 34 * the probability of getting a can that weighs 8 or more ounces. mean = n * p is a rule for binomial distributions. the standard deviation is sqrt(n * p * (1-p)). This is another rule for binomial distributions.

//

This would result in normalcdf(34 * 0.48 , 34 , 34 * 0.428 , sqrt(34 * 0.428 * (1-0.428))) = 0.269 <-- this is the answer