Question 1171229: Suppose a company repacks cereal with the use of a machine that is calibrated so that the weight of cereal dispensed in a box is normally distributed with mean 0.40 kg and variance 0.01 kg2, although the weight stated in the cereal box is 0.45 kg. For quality control purposes, suppose a random sample of 15 cereal boxes is taken. What is the probability of selecting a random sample with mean weight less than the advertised weight?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $X$ be the weight of cereal dispensed in a box. We are given that $X$ is normally distributed with mean $\mu = 0.40$ kg and variance $\sigma^2 = 0.01$ kg$^2$. Thus, the standard deviation is $\sigma = \sqrt{0.01} = 0.1$ kg.
We take a random sample of $n = 15$ cereal boxes. Let $\bar{X}$ be the sample mean. We want to find the probability that the sample mean is less than the advertised weight, which is 0.45 kg.
The sample mean $\bar{X}$ is also normally distributed with mean $\mu_{\bar{X}} = \mu = 0.40$ kg and variance $\sigma_{\bar{X}}^2 = \frac{\sigma^2}{n} = \frac{0.01}{15}$.
The standard deviation of the sample mean is $\sigma_{\bar{X}} = \sqrt{\frac{0.01}{15}} = \frac{0.1}{\sqrt{15}} \approx 0.02582$ kg.
We want to find $P(\bar{X} < 0.45)$. We standardize the sample mean using the z-score formula:
$$Z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}}$$
We calculate the z-score for $\bar{X} = 0.45$:
$$Z = \frac{0.45 - 0.40}{0.1/\sqrt{15}} = \frac{0.05}{0.1/\sqrt{15}} = \frac{0.05 \sqrt{15}}{0.1} = 0.5 \sqrt{15} \approx 1.9365$$
We want to find $P(Z < 1.9365)$. Using a standard normal distribution table or a calculator, we find:
$$P(Z < 1.9365) \approx 0.9736$$
Therefore, the probability of selecting a random sample with mean weight less than the advertised weight is approximately 0.9736.
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