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lord [1]
3 years ago
9

he amount of corn chips dispensed into a 12​-ounce bag by the dispensing machine has been identified as possessing a normal dist

ribution 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.

Mathematics
1 answer:
const2013 [10]3 years ago
3 0

Answer:

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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Sally is standing at one corner of a snowy rectangular field that measures 200 ft by 200 ft. She wishes toreach a warm cabin loc
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The optimal point(i.e the position of  P) that will allow Sally to reach the cabin as quickly as possible is  d =  50 \ ft from the cabin.

Step-by-step explanation:

From the question we are told that

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Let denote the distance from the bottom left corner to the  P  be  y  ft

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8 0
2 years ago
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