Question

The mean weight of chicken in a chicken dinner at a fast food restraint is 10 ounces with standard deviation of 0.5 ounces. Using the distribution of sample means, what is the probability that the average chicken weight in a sample of 100 dinners will differ from the mean by more than 0.03 ounces?

Answer 1

Answer on Question #44202 – Math - Statistics and Probability

Problem

The mean weight of chicken in a chicken dinner at a fast food restraint is 10 ounces with standard deviation of 0.5 ounces. Using the distribution of sample means, what is the probability that the average chicken weight in a sample of 100 dinners will differ from the mean by more than 0.03 ounces?

Solution

The weight of chicken is a random variable with mean $\mu = 10$ and standard deviation $\sigma = 0.5$. By central limit theorem, whatever the population, the distribution of $\bar{X}_n = \frac{X_1 + X_2 + \cdots + X_n}{n}$ is approximately normal when $n = 100$ is large.

It is known the average chicken weight in a sample of $n = 100$ dinners is a random variable with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$.

$$\bar{X}_n = \frac{X_1 + X_2 + \cdots + X_n}{n} \sim N\left(\mu, \frac{\sigma}{\sqrt{n}}\right),$$

i.e. $\bar{X}_{100} \sim N(10,0.05)$.

Consequently, the next variable is a standard normal with mean 0 and standard deviation 1:

$$Z = \frac{\bar{X}_{100} - E(\bar{X}_{100})}{\sqrt{D(\bar{X}_{100})}} = \frac{\bar{X}_{100} - \mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1).$$

Calculate

$$\begin{aligned} P(\bar{X}_{100} - 10 > 0.03) &= P(\bar{X}_{100} > 10.03) = 1 - P(\bar{X}_{100} \leq 10.03) = \\ &= 1 - P\left(\frac{\bar{X}_{100} - E(\bar{X}_{100})}{\sqrt{D(\bar{X}_{100})}} \leq \frac{10.03 - E(\bar{X}_{100})}{\sqrt{D(\bar{X}_{100})}}\right) = \\ &= 1 - P\left(\frac{\bar{X}_{100} - E(\bar{X}_{100})}{\sqrt{D(\bar{X}_{100})}} \leq \frac{10.03 - 10}{\frac{0.5}{10}}\right) \approx 1 - P\left(Z \leq \frac{0.03}{0.05}\right) = \\ &= 1 - P(Z \leq 0.6) = 1 - 0.7257 = 0.2743; \end{aligned}$$ $$\begin{aligned} P(\bar{X}_{100} - 10 < -0.03) &= P(\bar{X}_{100} < 9.97) = P\left(\frac{\bar{X}_{100} - E(\bar{X}_{100})}{\sqrt{D(\bar{X}_{100})}} < \frac{9.97 - E(\bar{X}_{100})}{\sqrt{D(\bar{X}_{100})}}\right) = \\ &= P\left(\frac{\bar{X}_{100} - E(\bar{X}_{100})}{\sqrt{D(\bar{X}_{100})}} < \frac{9.97 - 10}{\frac{0.5}{10}}\right) \approx P\left(Z \leq \frac{-0.03}{0.05}\right) = P(Z \leq -0.6) = \\ &= 0.2743 \end{aligned}$$

By additive property of probability, taking into account events $(\bar{X}_{100} - 10 > 0.03)$ and $(\bar{X}_{100} - 10 < -0.03)$ are mutually exclusive, therefore,

$$\begin{aligned} P(|\bar{X}_{100} - 10| > 0.03) &= P\left((\bar{X}_{100} - 10 > 0.03) \cup (\bar{X}_{100} - 10 < -0.03)\right) = \\ &= P(\bar{X}_{100} - 10 > 0.03) + P(\bar{X}_{100} - 10 < -0.03) = 0.2743 + 0.2743 = 0.5486 \approx 0.549. \end{aligned}$$

Answer. 0.549.

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