Question

The amount of corn chips dispensed into a 10-ounce bag by the dispensing machine has been identified aspossessing a normal distribution with a mean of 10.5 ounces and a standard deviation of 0.1 ounce. Suppose400 bags of chips were randomly selected from this dispensing machine. Find the probability that the sample mean weight of these 400 bags exceeded 10.6 ounces.

Answer 1

Answer:

The probability that the sample mean weight of these 400 bags exceeded 10.6 ounces is P(Xs>10.6)=0.

Step-by-step explanation:

When we take samples of size n=400, we have the folllowing parameters for the sampling distribution for the sample means:

$\mu_s=\mu=10.5\\\\ \sigma_s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{0.1}{\sqrt{400}}=\dfrac{0.1}{20}=0.005$

We can calculate the probability that the sample mean weight of these 400 bags exceeded 10.6 ounces calculating the z-score for Xs=10.6 and then its probability P(Xx>10.6), using the standard normal distribution:

$z=\dfrac{X_s-\mu_s}{\sigma_s}=\dfrac{10.6-10.5}{0.005}=\dfrac{0.1}{0.005}=20\\\\\\P(X_s>10.6)=P(z>20)=0$