If 16 oz of granola cost 5.99 how much does 1 oz of granola cost?
1 answer:
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Answer:
0.0322; 0.9929
Step-by-step explanation:
Since the data is normally distributed, we use z scores for these probabilities.
The formula for a z score of a sample mean is
For the sample of mildly obese people, the mean, μ, is 371; the standard deviation, σ, is 65; and the sample size, n, is 6.
Using 420 for X,
z = (420-371)/(65÷√6) = 49/(65÷2.4495) = 49/26.5360 ≈ 1.85
Using a z table, we see that the area under the curve to the left of this is 0.9678. However, we want the area to the right, so we subtract from 1:
1-0.9678 = 0.0322
For the sample of lean people, the mean, μ, is 528; the standard deviation, σ, is 108; the sample size, n, is 6.
Using 420 for X, we have
z = (420-528)/(108÷√6) = -108/(108÷2.4495) = -108/44.0906 ≈ -2.45
Using a z table, we see that the area under the curve to the left of this is 0.0071. We want the area under the curve to the right, so we subtract from 1:
1-0.0071 = 0.9929