# Question d6ec5

Sep 3, 2017

The distribution is not symmetric, centered at about 2-3, has a spread from 0-8 servings, and has no outliers. Percentages are below.

#### Explanation:

If you look at the left half of the graph and the right half of the graph, they don't look the same, so it is not symmetric.
The data looks like there are:

• 15 girls who ate 0 servings
• 11 girls who ate 1 serving
• 15 girls who ate 2 servings
• 11 who ate 3 servings
• 8 who ate 4 servings
• 5 who ate 5 servings
• 3 who ate each 6, 7, and 8 servings

Therefore, the number of girls involved in the study does add up to 74, and the middle, or median value is the 37th value from either end. If you were to write out the data, with fifteen $0$'s, eleven $1$'s, fifteen $2$'s, and so on, the 37th value would be a $2$. Likewise, the average, or mean, value is about $2.6$. So we say the center of the data is around 2 or 3 servings of fruit.
We say the data is skewed right because the graph has a "tail" going to the right. Google "skew of data" to see more examples.

The spread is the range whatever you're measuring, so here your spread is 0-8 pieces of fruit.
There are no outliers because all the data on the graph is grouped together. If there were some people, for example, who ate 19 pieces of fruit, that would be an outlier. Outliers are significant primarily because they affect the way our measures look. While they likely do not affect the median so much, they significantly change the mean value.

Percentages:
What percent ate 7 or more servings? Six of the girls ate 7 or more servings, and $\left(\frac{6}{74}\right) 100 \approx 2.6$. I don't know what "round to the nearest percent" instruction means. Maybe your instructor wants you to round to the nearest whole number, so I think 3% is a safe estimate.

For the percent who ate fewer than 3 servings, include
- 15 girls who ate 0 servings
- 11 girls who ate 1 serving
- 15 girls who ate 2 servings
...which is 41 of the participants.
(41/74)100~~55%#