Question

1. An impartial judge of a local garden competition collects scores from two groups of judges on what he suspects to be the number one garden. He finds the first group (n = 5) has a mean score of 76. The second group (n = 3) has a mean score of 91. So that he may declare the final score, what is the weighted mean of these two groups?

2. A social scientist measures the number of minutes (per day) that a small hypothetical population of college students spends online. Student Score Student Score A 94 F 96 B 88 G 25 C 74 H 61 D 88 I 82 E 98 J 98

(a) What is the range of data in this population? min

(b) What is the IQR of data in this population? min

(c) What is the SIQR of data in this population? min

(d) What is the population variance?

(e) What is the population standard deviation? (Round your answer to two decimal places.) min

3. A sociologist records the annual household income (in thousands of dollars) among a sample of families living in a high-crime neighborhood. Locate the lower, median, and upper quartiles for the times listed below. Hint: First arrange the data in numerical order.

a) lower quartile thousand dollars

b) median thousand dollars

c) upper quartile thousand dollars

42 22 46 33 37 32 37 47 51 25

4. A theme park owner records the number of times the same kids from two separate age groups ride the newest attraction. Age 13–16 Time Age 17–21 Time 1 10 1 4 2 9 2 3 3 3 3 7 4 1 4 4 5 10 5 8 6 3 6 1 7 8 7 2 8 9 8 5 9 6 9 4 10 5 10 2 Using the computational formula, for the age group of 13–16? (Round your answers for variance and standard deviation to two decimal places.) what is the

a) SS

b) sample variance

c) standard deviation

Answer 1

1. Weighted mean of these two groups=(5*76+3*91)/(5+3)=81.625

2.

(a) Minimum value=25, maximum value=98, then Range=73

(b) Arrange the data increasing order:

25, 61, 74, 82, 88, 88, 94, 96, 98, 98

Here N=total no. of observations=10

Now, (N+1)/4=2.75

First quartile=Q1=2nd observation of above ordered arrangement +0.75*(3rd observation of above ordered arrangement-2nd observation of above ordered arrangement)

=61+0.75*(74-61)=70.75

Now, 3(N+1)/4=8.25

Third quartile=Q3=8th observation of above ordered arrangement +0.25*(9th observation of above ordered arrangement-8th observation of above ordered arrangement)

=96+0.25*(98-96)=96.5

IQR=Q3-Q1=96.5-70.75=25.75

(c)

SIQR=(Q3-Q1)/2 =25.75/2=12.875

3.

Arrange the data increasing order:

22, 25, 32, 33, 37, 37, 42, 46, 47, 51

Here N=total no. of observations=10

a. Now, (N+1)/4=2.75

First quartile=Q1=2nd observation of above ordered arrangement +0.75*(3rd observation of above ordered arrangement-2nd observation of above ordered arrangement)

=25+0.75*(32-25)=30.25

b. Now 2*(N+1)/4=5.5

Median=Q2=5th observation of above ordered arrangement +0.50*(6th observation of above ordered arrangement-5th observation of above ordered arrangement)

=37+0.5*(37-37)=37

c. Now, 3(N+1)/4=8.25

Third quartile=Q3=8th observation of above ordered arrangement +0.25*(9th observation of above ordered arrangement-8th observation of above ordered arrangement)

=46+0.25*(47-46)=46.25