The probability that the sample mean weight of 36 randomly selected 10-ounce bags of corn chips is less than 10.20 ounces is approximately 0.9982, or 99.82%.
To find the probability that the sample mean weight of 36 randomly selected bags is less than 10.20 ounces, you can use the central limit theorem. According to the central limit theorem, if you have a sufficiently large sample size, the distribution of the sample mean will be approximately normal.
The standard deviation of the sample mean
is given by the population standard deviation
divided by the square root of the sample size n:
![\[ \sigma_{\bar{x}} = (\sigma)/(√(n)) \]](https://img.qammunity.org/2024/formulas/mathematics/college/p95n296w0k572nhz656ywjxd0m7s6sdxmi.png)
In this case,
(population standard deviation) is 0.4 ounces, and n (sample size) is 36. Therefore,
![\[ \sigma_{\bar{x}} = (0.4)/(√(36)) = (0.4)/(6) = 0.0667 \]](https://img.qammunity.org/2024/formulas/mathematics/college/ya23yijk8m9bkbx1lku6ev07ex6q5dgshe.png)
Now, you need to find the z-score for the sample mean of 10.20 ounces. The z-score is calculated using the formula:
![\[ z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} \]](https://img.qammunity.org/2024/formulas/mathematics/college/ft5ae96m02f7yhrp4jmxibbf5fbuq7poel.png)
where:
-
is the sample mean,
-
is the population mean,
-
is the standard deviation of the sample mean.
In this case,
, and
. Substituting these values:
![\[ z = (10.20 - 10)/(0.0667) \]\[ z \approx (0.20)/(0.0667) \]\[ z \approx 2.999 \]](https://img.qammunity.org/2024/formulas/mathematics/college/1bxn3s8mwj015dqucn47vifwlk1no9hfsn.png)
Now, you can use a standard normal distribution table or a calculator to find the probability that the z-score is less than 2.999.
The probability that the sample mean weight of 36 randomly selected bags is less than 10.20 ounces is the probability corresponding to a z-score of 2.999 in the standard normal distribution. The exact probability value can be looked up in a standard normal distribution table or calculated using a calculator.