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Part 1

Ms. MaiLam analyzed scores of 60 students, finding a mean of 3.15, median of 3, mode of 3, range of 5, variance of 2.028, and a standard deviation of 1.424, with a 95% confidence interval for average scores between 2.78 and 3.51. The manager of SNOWY RESTAURANT assessed wait times for 18 customers, resulting in a median wait time of 36 minutes, with a distribution of wait times across various intervals. The study highlighted customer concerns regarding long wait times, prompting the analysis.

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1 views3 pages

Part 1

Ms. MaiLam analyzed scores of 60 students, finding a mean of 3.15, median of 3, mode of 3, range of 5, variance of 2.028, and a standard deviation of 1.424, with a 95% confidence interval for average scores between 2.78 and 3.51. The manager of SNOWY RESTAURANT assessed wait times for 18 customers, resulting in a median wait time of 36 minutes, with a distribution of wait times across various intervals. The study highlighted customer concerns regarding long wait times, prompting the analysis.

Uploaded by

phuloi1239
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PART 1:

QUESTION 1:

Ms. MaiLam (A teacher) selected a random sample of 60 students at H and recorded scores
of these students. The following data were recorded and summarized in table below:

Score Frequency (number of students)


0 3
1 6
2 8
3 17
4 14
5 12

∑ x ( 0 x 3 ) + ( 1 x 6 )+ ( 2 x 8 ) + ( 3 x 17 )+ ( 4 x 14 ) +(5 x 12)
a. Mean: x = = = 3.15
n 60
n 60
Median: i= = = 30th (is an integer) => i+1= 31 => at 31th (3) => i = 3
2 2
Mode: 3 (the score with highest frequency)
Range: R= Max - Min = 5 – 0 = 5
2
∑ ( x−x ) 119.65
Variance: s2 = = (0−3.15)2 x 3+(1−3.15)2 x 6+ …+¿ ¿ = =2.028
n−1 59
Standard deviation: s = √ s 2 = √ 2.028=1.424
s 1.424
Coefficient of variation (CV): CV= x 100 = x100 = 45.21%
x 3.15
b. We know confident level is 95%, n=60 but we don’t know population standard =>
we use Z.
 Z = 1.96

We have

s 1.424
x ± z x = 3.15 ± 1.96 x => 2.78 < M < 3.51
√n √ 60
With 95% confident level, the averages scores of all students at H is from 2.78 to
3.51.

QUESTION 2:
The manager of the SNOWY RESTAURANT recently selected a random sample of 18
customers and kept track of how long the customers were required to wait from the time
they arrived at the restaurant until they were actually served dinner. This study resulted
from several complaints the manager had received from customers saying that their wait
time was unduly long and that it appeared that the objective was to keep people waiting in
the lounge for as long as possible to increase the lounge business. The following data were
recorded, with time measured in minutes (use as the continuous data):

34 24 43 56 74 20 19 33 55 43 54 34 27 34 36 24 54 39

a. In order, we have:

19 20 24 24 27 33 34 34 34 36 39 43 43 54 54 55 56 74

We have:

 K=6
 H = 10
 The lower limit of the first class 15
 Min = 19
 Max = 74

Distribution table:

Data Frequency Cumulative Frequency Relative Frequency (%)


15 – 24 4 4 22.2%
25 – 34 5 9 27.7%
35 – 44 4 13 22.2%
45 – 54 2 15 11.1%
55 – 64 2 17 11.1%
65 – 74 1 18 5.5%
N = 18 Total: 100%

n 18
b. Median: = = 9 (is an integer) => i+1 = 10
2 2
Median will be in 10th value (36)
n 18
Q1: i = = = 4.5 (is not an integer) => 5
4 4
Q1 will be in 5th value (27)
3 3
Q3: i = n = x 18 = 13.5 (is not an integer) => 14
4 4
Q3 will be in 14th value (54)
Lower limit = Q1 – 1.5 x (Q3 – Q1) = -13.5
Upper limit = Q3 + 1.5 x (Q3 – Q1) = 94.5

Median
Q1 Q3
-13.5 0 94.5

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