The Acme Candy Company claims that 60% of the jawbreakers it produces weigh more than 0.4 ounces. Suppose that 800 jawbreakers are selected at random from the production lines. Would it be unusual for this sample of 800 to contain 470 jawbreakers that weigh more than 0.4 ounces? Consider as unusual any result that differs from the mean by more than 2 standard deviations.
+169 I'm going to assume the normal distribution applies to this problem.
$$P=\int_{lb}^{rb}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\mathrb{d}x$$
With P being the chance that a certain amount of jawbreaker has a weight between $$lb$$ and $$rb$$, respectively being the left bound and the right bound. In this equation $$\mu$$ is the mean and $$\sigma$$ the standarddeviation.
At this time we have no knowledge of either $$\mu$$ or $$\sigma$$, all we know for sure is that:
$$P(weight>0.4)=0.60$$
Hence the left bound is equal to 0.4 and the right bound to infinity. We get the equation:
$$P(weight>0.4)=\int_{0.4}^{\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\mathrb{d}x=0.60$$
This still leaves us with two unknown variables: $$\mu$$ and $$\sigma$$.
When analyzing the function we gain the knowledge that when $$\sigma$$ approaches zero, $$\mu$$ approaches the value 0.4. This means:
$$\lim_{\sigma\to0}\Rightarrow\lim_{\mu\to0.4$$
Since the example shows a different outcome to P(weight>0.4):
$$P(weight>0.4)=\frac{470}{800}\approx0.59$$
We can let $$\sigma$$ approach zero and render the outcome unusual. In a similar manner we can pick $$\sigma$$ in such a way that the sample is in line with the distribution.
In short, the question doesn't provide enough information to say for certain if the sample is unusual or not.
+169 I'm going to assume the normal distribution applies to this problem.
$$P=\int_{lb}^{rb}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\mathrb{d}x$$
With P being the chance that a certain amount of jawbreaker has a weight between $$lb$$ and $$rb$$, respectively being the left bound and the right bound. In this equation $$\mu$$ is the mean and $$\sigma$$ the standarddeviation.
At this time we have no knowledge of either $$\mu$$ or $$\sigma$$, all we know for sure is that:
$$P(weight>0.4)=0.60$$
Hence the left bound is equal to 0.4 and the right bound to infinity. We get the equation:
$$P(weight>0.4)=\int_{0.4}^{\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\mathrb{d}x=0.60$$
This still leaves us with two unknown variables: $$\mu$$ and $$\sigma$$.
When analyzing the function we gain the knowledge that when $$\sigma$$ approaches zero, $$\mu$$ approaches the value 0.4. This means:
$$\lim_{\sigma\to0}\Rightarrow\lim_{\mu\to0.4$$
Since the example shows a different outcome to P(weight>0.4):
$$P(weight>0.4)=\frac{470}{800}\approx0.59$$
We can let $$\sigma$$ approach zero and render the outcome unusual. In a similar manner we can pick $$\sigma$$ in such a way that the sample is in line with the distribution.
In short, the question doesn't provide enough information to say for certain if the sample is unusual or not.
