The infant mortality rate is defined as the number of infant deaths per 1,000 live births. this rate is often used as an indicator of the level of health in a country. the relative frequency histogram below shows the distribution of estimated infant death rates for 224 countries for which such data were available in 2014.31
(a) estimate q1, the median, and q3 from the histogram.
(b) would you expect the mean of this data set to be smaller or larger than the median? explain your reasoning.
the graph that goes with it is in the picture.

Here, we are required to estimate Q1, the median, and Q3 from the histogram and determine if the mean of this data set is smaller or larger than the median.

(a)

Q1 = 7

Q1 = 7median = 15

Q1 = 7median = 15Q3 = 40

(b). When data plotted on a bar chart as this, is heavily skewed to the left, the mean is very much larger than the median.

(a) To estimate first quartile, Q1, median (Q2) and third quartile, Q3.

1. The first quartile corresponds to 1/4 i.e 0.25 of the cumulative frequency distribution. It's in the first bar as the bar corresponds to 0.38 on the fraction of countries axis. Therefore, we can estimate Q1 = 5.

However, since 0.38 is slightly higher than 0.25, we can estimate Q1 to be slightly higher than 5.

And since the bar represents a range from 0 to 10

Therefore, Q1 = 7

2. Secondly, to estimate the median, the point where the 2nd quartile falls on the cumulative frequency distribution (i.e the 0.5 fraction).

Therefore, since the 1st bar has 0.38 fraction of the frequency, and the 2nd bar has 0.22 of the frequency. The total of both bars gives a cumulative frequency of 0.60.

Therefore, since the median falls around half way through the second bar, the median can be estimated as median = 15.

3. Lastly, Q3 falls around the 0.75 of the cumulative frequency distribution.

Since the third bar corresponds to 0.11 fraction of countries and the 4th bar corresponds to 0.07 fraction of countries.

Consequently, the cumulative frequency of the fraction of countries becomes 0.60 + 0.11 + 0.05.

Total then becomes 0.76.

The Q3 is therefore around the far end of the 4th bar and then can be estimated as Q3 = 40.

(b). When data plotted on a bar chart as this, is heavily skewed to the left, the mean is very much larger than the median.

By observing, the frequency of data from the far end. The cumulative frequency of the data from 80 to 120 falls around 5% so adds only a small impact on the mean, same for the cumulative frequency of the data from 80 to 60 and 60 to 40. Therefore, the mean is then way larger than the median.

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(a)

Q1, the first quartile, 25th percentile, is greater than or equal to 1/4 of the points.  It's in the first bar so we can estimate Q1=5.   In reality the bar includes values from 0 to 9 or 10  (not clear which) and has around 37% of the points so we might estimate Q1 a bit higher as it's 2/3 of the points, say Q1=7.

The median is bigger than half the points.  First bar is 37%, next is 22%, so its about halfway in the second bar, median=15

Third bar is 11%, so 70% so far.  Four bar is 5%, so we're at the right end of the fourth bar for Q3, the third quartile, 75th percentile, say Q3=40

b

When the data is heavily skewed left like it is here, the median tends to be lower than the mean.  The 5% of the data from 80 to 120 averages around 100 so adds 5 to the mean, and 8% of the data from the 60 to 80 adds another 5.6, 15% of the data from 40 to 60 adds about 7.5, plus the rest, so the mean is gonna be way bigger than the median of around 15.

$53.50

step-by-step explanation:

you have to subtract the amount of money chuck earned ($72.50) by the cost of the gas for the mower ($11.25) and the new string for the trimmer he bought ($7.75).

72.50-11.25=61.25

and then subtract $61.25 by the second thing he bought (the new string for the trimmer)

61.25-7.75=53.5

so in total it's $53.50 (53 dollars and 50 cents)

a and c

step-by-step explanation: