Question

Use 5 decimal places for intermediate calculations, and round your final answers to 3 decimal places.

The weights of a random sample of pet food bags that are supposed to weigh 1010 pounds are given below. Assume the population is normally distributed.

9.6 10.4 10.2 9.5 10.5 10

(a) Find the sample mean.

x¯=

(b) Find the sample variance and sample standard deviation.

s2= and s=

(c) Find a 98% confidence interval for the mean weight of the bags.

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Answer 1

The sample mean is approximately 10.033, the sample variance is approximately 0.124, and the sample standard deviation is approximately 0.351.

Step-by-step explanation:

To find the sample mean, we need to sum up all the weights and divide by the number of observations. The weights given are 9.6, 10.4, 10.2, 9.5, 10.5, and 10. Let's calculate the sample mean:

Sample Mean:

(9.6 + 10.4 + 10.2 + 9.5 + 10.5 + 10) / 6 = 60.2 / 6 = 10.0333

Rounding to 3 decimal places, the sample mean is approximately 10.033.

To find the sample variance, we need to calculate the squared deviations from the sample mean and find their average. Let's calculate the sample variance:

Sample Variance:

((9.6 - 10.0333)^2 + (10.4 - 10.0333)^2 + (10.2 - 10.0333)^2 + (9.5 - 10.0333)^2 + (10.5 - 10.0333)^2 + (10 - 10.0333)^2) / 6 = 0.12378333

Rounding to 3 decimal places, the sample variance is approximately 0.124.

Finally, to find the sample standard deviation, we take the square root of the sample variance. Let's calculate the sample standard deviation:

Sample Standard Deviation:

sqrt(0.124) = 0.351363

Rounding to 3 decimal places, the sample standard deviation is approximately 0.351.