The following table shows the number of hours some middle school students in two cities spend texting each week: City A 9 27 17
24 21 12 25 25 18 City B 6 20 26 15 23 25 14 14 11 Part A: Create a five-number summary and calculate the interquartile range for the two sets of data. (6 points) Part B: Are the box plots symmetric? Justify your answer. (4 points)
1 answer:
Five-number summaries are: lowest value, highest value, upper quartile (Q₃), median (Q₂) and lower quartile (Q₁)
The first step is to rearrange each data set in ascending order as shown in the first diagram
The median (Q₂) of each data set is located on the
value.
Once the median is located, we then have the set of data on the left and on the right of median where we are going to locate the Q₁ and Q₃.
Q₁ is located in between the
and the 
value and Q₃ is located between the 
and the 
value.
The box plot of each data set are shown in the second diagram
The box plots aren't symmetric. Box plot for City A tails on the left (negative skew) and box plot for city B also tails on the left (negative skew).
The first step is to rearrange each data set in ascending order as shown in the first diagram
The median (Q₂) of each data set is located on the
Once the median is located, we then have the set of data on the left and on the right of median where we are going to locate the Q₁ and Q₃.
Q₁ is located in between the
The box plot of each data set are shown in the second diagram
The box plots aren't symmetric. Box plot for City A tails on the left (negative skew) and box plot for city B also tails on the left (negative skew).
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The probability of picking one girl would be . That is because there are 5 girls out of the 12 students, and the probability of an event occuring is:
.
Using that same logic, the next student should be easier. We reduced the student population by 1, so we have 11 possible ways it can happen now instead of 12, so that gives us:, for the probability of picking a boy as the second pick.
And lastly, using the same logic shown above, the probability of picking a girl on the third pick would be:.
We are not done, though. We have the separate probabilities, but now we have to multiply then together to figure out the probability of this exact event happening:
Which when reduced is:
Answer:
The 95% of confidence intervals
(2.84 ,2.99)
Step-by-step explanation:
A random sample of 20 accounting students results in a mean of 2.92 and a standard deviation of 0.16
given small sample size n =20
sample mean x⁻ =2.92
sample standard deviation 'S' =0.16
level of significance ∝ = 0.95
The 95% of confidence intervals
the degrees of freedom γ=n-1 =20-1=19
t-table 2.093
(2.92-0.0748,2.92+0.0748)
(2.84 ,2.99)
Therefore the 95% of confidence intervals
(2.84 ,2.99)
