(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.
-87 is the correct answer
Answer:
43
Step-by-step explanation:
Using the z-distribution, as we have the standard deviation for the population, it is found that the smallest sample size required to obtain the desired margin of error is of 77.
<h3>What is a z-distribution confidence interval?</h3>
The confidence interval is:
In which:
is the sample mean.
is the standard deviation for the population.
The margin of error is given by:
In this problem, we have that the parameters are given as follows:
.
Hence, solving for n, we find the sample size.
Rounding up, the smallest sample size required to obtain the desired margin of error is of 77.
More can be learned about the z-distribution at brainly.com/question/25890103
