Question

An article reported that the mean annual adult consumption of wine was 3.85 gallons and that the standard deviation was 6.07 gallons. Would you use the empirical rule to approximate the proportion of adults who consume more than 9.92 gallons (i.E., the proportion of adults whose consumption value exceeds the mean by more than 1 standard deviation)? Explain your reasoning. (Round your numerical answer to three decimal places.)

Answer 1

Answer:

0.159 is the required proportion.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 3.85 gallons

Standard Deviation, σ = 6.07 gallons

We are given that the distribution of consumption of wine is a bell shaped distribution that is a normal distribution.

Empirical Formula:

• Almost all the data lies within three standard deviation from the mean for a normally distributed data.
• About 68.2% of data lies within one standard deviation from the mean.
• About 95% of data lies within two standard deviations of the mean.
• About 99.7% of data lies within three standard deviation of the mean.

We have to evaluate:

$=P(x \geq 9.92)\\=P(x \geq \mu + 1(\sigma))\\=0.5 - \dfrac{1}{2}(0.68.2)\\\\=0.5 - 0.34\\=0.159$

0.159 is the proportion of adults whose consumption value exceeds the mean by more than 1 standard deviation.