The mean cost of a five pound bag of shrimp is 50 dollars with a standard deviation of 8 dollars.
If a sample of 56 bags of shrimp is randomly selected, what is the probability that the sample mean would differ from the true mean by less than
0.9 dollars? Round your answer to four decimal places.
AnswerHow to enter your onswer (opens in new window) 2 points

Question

UpStudy AI · Accepted Answer

To find the probability that the sample mean differs from the true mean by less than $0.9, we can use the Central Limit Theorem. According to the theorem, the distribution of the sample mean will be approximately normally distributed if the sample size is sufficiently large (which it is in this case, with a sample size of 56).

### Step 1: Calculate the standard error of the mean (SEM)

The standard error of the mean is calculated using the formula:

$$
\text{SEM} = \frac{\sigma}{\sqrt{n}}
$$

where:
- $$\sigma$$ is the population standard deviation (8 dollars),
- $$n$$ is the sample size (56).

$$
\text{SEM} = \frac{8}{\sqrt{56}} \approx \frac{8}{7.4833} \approx 1.0672
$$

### Step 2: Calculate the z-scores for the sample mean

We want to find the probability that the sample mean differs from the true mean by less than $0.9. This means we are looking for:

$$
P(-0.9 < \bar{x} - 50 < 0.9)
$$

This can be rewritten in terms of z-scores:

$$
P(50 - 0.9 < \bar{x} < 50 + 0.9) = P(49.1 < \bar{x} < 50.9)
$$

Now we convert these values to z-scores:

$$
z = \frac{\bar{x} - \mu}{\text{SEM}}
$$

For $$49.1$$:

$$
z_1 = \frac{49.1 - 50}{1.0672} \approx \frac{-0.9}{1.0672} \approx -0.843
$$

For $$50.9$$:

$$
z_2 = \frac{50.9 - 50}{1.0672} \approx \frac{0.9}{1.0672} \approx 0.843
$$

### Step 3: Find the probabilities corresponding to the z-scores

Using the standard normal distribution table or a calculator:

- For $$z_1 \approx -0.843$$, the cumulative probability $$P(Z < -0.843) \approx 0.1997$$.
- For $$z_2 \approx 0.843$$, the cumulative probability $$P(Z < 0.843) \approx 0.7997$$.

### Step 4: Calculate the probability that the sample mean is within the range

Now, we find the probability that the sample mean is between these two z-scores:

$$
P(-0.843 < Z < 0.843) = P(Z < 0.843) - P(Z < -0.843)
$$

$$
P(-0.843 < Z < 0.843) \approx 0.7997 - 0.1997 = 0.6000
$$

### Final Answer

Thus, the probability that the sample mean would differ from the true mean by less than $0.9 is approximately:

$$
\boxed{0.6000}
$$
The probability is approximately 0.6000.

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Statistics Questions