Uploaded by Denise Jane Roque

-STT-0411-Quiz-4-converted

advertisement
ROQUE, DENISE JANE G.
STT0411 G5.13
Quiz #4
I.
Determine whether the numerical value is a parameter or a statistic. (4)
1. The 2009 team payroll of the XYZ company was Php109, 004, 046. Parameter
2. In a survey of 1052 adults in the Philippines, 39% think there should be a
law that prohibits people from talking on cell phones in public places. Statistics
Identify the sampling technique used in each study. (6)
3. Using random digit dialing, researchers ask 1003 U.S. adults their plans on
working during retirement. (Source: Princeton Survey Research
Associates International) – Simple Random Sampling
4. A student asks 18 friends to participate in a psychology experiment.Convenience Sampling
5. A pregnancy study in Cebu, Philippines randomly selects 33 communities
from the Cebu metropolitan area, then interviews all available pregnant
women in these communities. (Adapted from Cebu Longitudinal Health
and Nutrition Survey) – Cluster Sampling
The data set represents the number of minutes a sample of 25 people exercise
each week.
108 139 120 123 120 132 123 131 131
157 150 124 111 101 135 119 116 117
127 128 139 119 118 114 127
6. Make a frequency distribution of the data set using five classes. Include
class limits, midpoints, boundaries, frequencies, relative frequencies, and
cumulative frequencies. (18pts)
II.
III.
Frequency Distribution Table:
Least Observation =101
Greatest Observation = 157
No. of classes = 5
Class range = (157-101)/5 = 11.2
Class Limit
Midpoint
Boundaries
Upper
Limit
Lower
Limit
Frequency
Relative Freq.
Cumulative Freq.
101-112
106.5
100.5
112.5
3
.12
3
113-123
118
112.5
123.5
10
.40
13
124-134
129
123.5
134.5
7
.28
20
135-145
140
134.5
145.5
3
.12
23
146-156
151
145.5
156.5
2
.08
25
∑ f = 25
f
∑ = 1.00
n
7.
Display the data using a frequency histogram. (3)
8. Describe the distribution’s shape. (2)
-The data distribution shown has a right skewed distribution, with the tail on the
right side.
IV.
Forty-three percent of businesses in the United States require a doctor’s note
when an employee takes sick time. You randomly select five businesses and ask
each if it requires a doctor’s note when an employee takes sick time.
9. Construct a binomial distribution (5)
p= .43 , q=.57, n = 5
P(0)= {(5C0) * (0.43^0) * (0.57^5)} = 0.060
P(1)={(5C1) * (0.43^1) * (0.57^4)} = 0.227
P(2) = {(5C2) * (0.43^2) * (0.57^3)} = 0.342
P(3)= {(5C3) * (0.43^3) * (0.57^2)} = 0.258
P(4) = {(5C4) * (0.43^4) * (0.57^1)} = 0.097
P(5) = {(5C5) * (0.43^5) * (0.57^0)} = 0.015
≈ (0.060 + 0.227 +0.342 0.258 + 0.097 + 0.015)
≈ 1.00
10. Find the mean, variance, and standard deviation of the binomial
distribution and interpret the results in the context of the real-life situation,
and (6)
Two out of every five employers require a doctor's letter when an
employee takes sick absence. Doctor's notes are not needed to be
followed up on because not all companies offer health and sick leave
incentives.
V.
11. Determine the value/s of the random variable x that you would
consider unusual. Explain why? (3)
When an employee takes sick time, according to the figures
above, it's rare for all five to require a doctor's letter
because not all corporate sectors have a sick leave
incentive that requires a doctor's note for sick time.
A survey indicates that for each trip to the supermarket, a shopper spends an
average of 50 minutes with a standard deviation of 13 minutes in the store. The
lengths of time spent in the store are normally distributed and are represented
by the variable x. A shopper enters the store.
12. Find the probability that the shopper will be in the store between 37 and 57
minutes. (5)
13. Interpret your answer if 125 shoppers enter the store. How many shoppers
would you expect to be in the store between 37 and 57 minutes?(3)
Between 37 and 57 minutes, 55 percent of shoppers will enter the
store. At that time, there will be (125 * 54.67 percent = 68. 34), or
around 68 shoppers in the store.
Download