He amount of corn chips dispensed into a 12-ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 12.5 ounces and a standard deviation of 0.2 ounces. Suppose 100 bags of chips were randomly selected from this dispensing machine. Find the probability that the sample mean weight of these 100 bags exceeded 12.6 ounces. Round to four decimal places.

The probability will be 0.3085 or 0

Step-by-step explanation:

Given:

True mean=12.5

Sample mean =12.6

Standard deviation=0.2

Samples=100

To Find:

Probability that exceeds 12.6 ounces.

Solution:

Calculate the Z-score for given means and standard deviation.

So

Z-score= (true mean -sample mean)/standard deviation.

Z-score=(12.5 -12.6)/0.2

=-0.1/0.2

=-0.5

Now Using Z-table

P(X≥-0.5)=p(Z≥-0.5)=0.3085

Hence Probability that sample mean weight exceeds will be  0.3085

                 OR

By using Normal distribution with sampling ,it will be as follows

Z=(X-u)/[Standard deviation/Sqrt(No of samples)]

Z=(12.6-12.5)/(0.2/Sqrt(100)

Z=0.1/0.2/10

Z=5

So P(X≥12.6  )=P(Z≥5)=1

Pr(Z≥5)=1-1=0.

(Refer the attachment )

Hence Probability of getting ounces greater than 12.6 is '0'.

The sampling is of 0.02 size hence graphically it looks likely.

as shown in attachment.

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step-by-step explanation:

the correct answers i think are b. and d.

give me brainliest if it is correct!