Mrs. Pellegrin has weighed 5 packages of cheese and recorded the weights as 10.2 oz, 10.5 oz, 9.3 oz, 9.8 oz, and 10.0 oz. She calculated the standard deviation to be 0.45 oz. Select the 95% confidence interval for Mrs. Pellegrin's set of data. a. 9.4 to 10.52.
b. 9.48 to 10.44.
c. 9.34 to 10.44.
d. 9.53 to 10.39.

The 95% Confidence Interval = (9.55, 10.35)

Step-by-step explanation:

Confidence Interval formula

=Mean ± z × standard deviation/√n

Step 1

We find the mean

Mean

10.2 oz + 10.5 oz + 9.3 oz + 9.8 oz + 10.0 oz/5

Mean = 49.8/5

Means = 9.96 0z

the standard deviation to be 0.45 oz. the z score for 95% confidence interval = 1.96

n = 5 samples

Hence, Confidence Interval =

9.96 ± 1.96 × 0.45/√5

9.96 ± 1.96 × 0.201246118

9.96 ± 0.3944423912

Confidence Interval =

9.96 - 0.3944423912

= 9.5655576088

≈ 9.55

9.96 + 0.3944423912

= 10.354442391

≈ 10.35

Therefore, the 95% Confidence Interval = (9.55, 10.35)

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