3. What concepts or properties do you need to solve the problems?

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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3. What concepts or properties do you need to solve the problems?
Transcribed Image Text:3. What concepts or properties do you need to solve the problems?
B. Read the given article/texts carefully. Then, complete the table and answer the questions that follow,
Essential
Text 1
Text 2
Text 3
Question
The computer terminals in a
school are labeled 1 to 300,
The technician wants to
know if the computers are in
good working condition by
testing 50 of these units. He
randomly selects the 12th to
be the first unit to be tested.
Which computer unit will be
tested next?
Fifty people were surveyed
The table shows the
on the number of full-length frequency count of the
employees' responses to the
survey of employees'
satisfaction level. A
numerical rating of 5 is
assigned to very satisfied
(VS), 4 to satisfied (S), 3 to
neutral (N), 2 to somewhat
dissatisfied (SD), and 1 to
very dissatisfied (VD).
Number of
Employces
25
20
movies watched in a year.
Number of
Регsons
Number of
Movies
Watched
19-21
16-18
1
13-15
5
7
10-12
7-9
8
4-6
0-3
At least how many movies
should one watch to belong
to upper 5th percentile?
12
15
Level of
Satisfaction
VS (5)
S (4)
N (3)
SD (2)
VD (1)
What is the employees' Level
of Satisfaction with the
13
4
3
How can
issues and
company? And at what
percent?
Answer:
problems on
your
respective
community
be solved?
Answer:
Answer:
Ore should wotch at legst 1625 or 17 movies to belog to The employees Level of Satisfaction with the company is The computer unit will be tested next is the 18th comouter
the upper 5th percentile.
3,92 or aprox, (Satisfied) with 78,4% satisfaction rate,
Supporting Texts:
Supporting Texts:
Supporting Texts:
A person should wateh at least 17 movies to belong to the
Based on the aample groupad data of 65 amployeea, the
The computer termnale in a school are labeled 1 to 300.
mean is 3.92 or approximately 4 which indioate as
upper Sth percentile or the 95th percentile. This is beoause Sutiefied (S) on the level of sutiefootion This le because I The techrikian wants to teet 50 urita and rardomly selecta
we used the formula Pru bpo +In*t/100 - of /fpe 1i
eome up to my own eomputation and formula which la
where bpen first olass boundary in the loated cumulative weighted means VS(25) + K(20) + NC13) + SDC4) +
VD) / total number of employees where VS= very
atiafind, S- autiufied, N- rentral, SD- whut
12h to be the first unit to be tested, The second or the
next computer urit will be tested is the 18th unit, This is
frequeney, ne value we need to find, fe frequeney of the
value (total number of persons), and of oumulative
because we use my buit up 2 formala which is umbere
of planned to test computer units total computer termhak
dissatisfied, VD= very dissatisfied. the mean is 3.92 or
approximately 4 which indicate as Saliefied (S) en the
level of sutisfuction, and total number of employeas 65,
frequeney of the closs before the class with the value of
n schools" and "first unit to be tested + anawer for the
the given perentile.
first formula".
Reason:
Reason:
First, we get the the cumalative frequency and class
Reason:
To determine the level of satisfaction, we simply get the
For this problem, we use my buit up 2 formulas and
bourdaries to oomplete the table, now we lve for the
mean by getting the sum of multiplied frequency to ita
percentile. So we reed to colesiate the 95th percentile. We numerical value and divided it by the total frequency form
the formula "weighted mean= VS(25) + S(20) + NC13) +
upper Sth percentile, Upper Sth percentile is 100-5-95th
aubstitute the values needed, There are 300 oomputer
terminale in a school Fifty units are needed to test if the
uoe the formula "Pr Ibpo +Entf/100 - ef fpa 11", We
computers are in good working condition, Since the first unit
to be tested is 12th, we need to divide firet the number of
eumulative frequeney volue greater or equal to 47.5. After substitute the given vales to the formua, weghted meane eomputer terminals which is 300 to the numbers of units
first find the n*T/100. 95*50 (50 ie from the total number
SD(4) + VD(1) / total number of employees". We
of persons)a 4,750/100- 47.5. Then we loeate the
dutermiring, we firnd the n-95, bde-15.5, et-47, fpe-2,
5(25) + 4(20) + 3(13) + 2(4) + 1(1) / 65 then we get the
needed to test which is 50, It is equal to 6 as the intervol
of testing, atter that we use the first unit to be tested
and i=18.5-15.5-3. We subetitute the values in the formula
answer of 3.92 and if we round it off, we'l get the anawer
then inturprut, P95- 15.5 (47,5- 47/ 2) 3, and the
answer is 16.25. Se vakue is 16.25 is in the upper 5th
-of 4 that shows the employees' level of satisfaction of
which is 12 to be added to the interval of testing and the
percentile or the 95th percentile round 16,25 up te the
nearest whole number is 17 Tie io because we oon't have
Satisfind, And to determine the level of satisfaction in
percent, we divided the mean to the highest value of level
sum is 18 which will be the next computer unit to be tested.
decimcal valu tor number of movina xit is a disernie
vartable). So we must have at least 17 movles to watch to of satisfaction and multiply it by 100 then we get the
helong tn the pper fith pereonrililn.
anawer of 78,4%
Common Ideas in Reasons:
It is important to note that the observation are collected randomly, hence,
there is no given biased/preference for one observation over another
> It's worth noting that the observations are gathered at random, so there's no
prejudice or biased for one observation over another.
Enduring Understanding/Generalization:
Frequeny couts, grouped data, serve as pood starting of desripitive
statistically analysis given a data set it gives more structured/ compact
analysis and it easinr /mure convenient to use as compared to ungrouped
data
> When oppused to ungroaped tlata, fruuency cuurta, BrOuped data, Jerved
as a uselul beginning point for deseriptive statiratically analysis given a data
het. It provideu a more structured/ condensed analysin and is uusier/more
Transcribed Image Text:B. Read the given article/texts carefully. Then, complete the table and answer the questions that follow, Essential Text 1 Text 2 Text 3 Question The computer terminals in a school are labeled 1 to 300, The technician wants to know if the computers are in good working condition by testing 50 of these units. He randomly selects the 12th to be the first unit to be tested. Which computer unit will be tested next? Fifty people were surveyed The table shows the on the number of full-length frequency count of the employees' responses to the survey of employees' satisfaction level. A numerical rating of 5 is assigned to very satisfied (VS), 4 to satisfied (S), 3 to neutral (N), 2 to somewhat dissatisfied (SD), and 1 to very dissatisfied (VD). Number of Employces 25 20 movies watched in a year. Number of Регsons Number of Movies Watched 19-21 16-18 1 13-15 5 7 10-12 7-9 8 4-6 0-3 At least how many movies should one watch to belong to upper 5th percentile? 12 15 Level of Satisfaction VS (5) S (4) N (3) SD (2) VD (1) What is the employees' Level of Satisfaction with the 13 4 3 How can issues and company? And at what percent? Answer: problems on your respective community be solved? Answer: Answer: Ore should wotch at legst 1625 or 17 movies to belog to The employees Level of Satisfaction with the company is The computer unit will be tested next is the 18th comouter the upper 5th percentile. 3,92 or aprox, (Satisfied) with 78,4% satisfaction rate, Supporting Texts: Supporting Texts: Supporting Texts: A person should wateh at least 17 movies to belong to the Based on the aample groupad data of 65 amployeea, the The computer termnale in a school are labeled 1 to 300. mean is 3.92 or approximately 4 which indioate as upper Sth percentile or the 95th percentile. This is beoause Sutiefied (S) on the level of sutiefootion This le because I The techrikian wants to teet 50 urita and rardomly selecta we used the formula Pru bpo +In*t/100 - of /fpe 1i eome up to my own eomputation and formula which la where bpen first olass boundary in the loated cumulative weighted means VS(25) + K(20) + NC13) + SDC4) + VD) / total number of employees where VS= very atiafind, S- autiufied, N- rentral, SD- whut 12h to be the first unit to be tested, The second or the next computer urit will be tested is the 18th unit, This is frequeney, ne value we need to find, fe frequeney of the value (total number of persons), and of oumulative because we use my buit up 2 formala which is umbere of planned to test computer units total computer termhak dissatisfied, VD= very dissatisfied. the mean is 3.92 or approximately 4 which indicate as Saliefied (S) en the level of sutisfuction, and total number of employeas 65, frequeney of the closs before the class with the value of n schools" and "first unit to be tested + anawer for the the given perentile. first formula". Reason: Reason: First, we get the the cumalative frequency and class Reason: To determine the level of satisfaction, we simply get the For this problem, we use my buit up 2 formulas and bourdaries to oomplete the table, now we lve for the mean by getting the sum of multiplied frequency to ita percentile. So we reed to colesiate the 95th percentile. We numerical value and divided it by the total frequency form the formula "weighted mean= VS(25) + S(20) + NC13) + upper Sth percentile, Upper Sth percentile is 100-5-95th aubstitute the values needed, There are 300 oomputer terminale in a school Fifty units are needed to test if the uoe the formula "Pr Ibpo +Entf/100 - ef fpa 11", We computers are in good working condition, Since the first unit to be tested is 12th, we need to divide firet the number of eumulative frequeney volue greater or equal to 47.5. After substitute the given vales to the formua, weghted meane eomputer terminals which is 300 to the numbers of units first find the n*T/100. 95*50 (50 ie from the total number SD(4) + VD(1) / total number of employees". We of persons)a 4,750/100- 47.5. Then we loeate the dutermiring, we firnd the n-95, bde-15.5, et-47, fpe-2, 5(25) + 4(20) + 3(13) + 2(4) + 1(1) / 65 then we get the needed to test which is 50, It is equal to 6 as the intervol of testing, atter that we use the first unit to be tested and i=18.5-15.5-3. We subetitute the values in the formula answer of 3.92 and if we round it off, we'l get the anawer then inturprut, P95- 15.5 (47,5- 47/ 2) 3, and the answer is 16.25. Se vakue is 16.25 is in the upper 5th -of 4 that shows the employees' level of satisfaction of which is 12 to be added to the interval of testing and the percentile or the 95th percentile round 16,25 up te the nearest whole number is 17 Tie io because we oon't have Satisfind, And to determine the level of satisfaction in percent, we divided the mean to the highest value of level sum is 18 which will be the next computer unit to be tested. decimcal valu tor number of movina xit is a disernie vartable). So we must have at least 17 movles to watch to of satisfaction and multiply it by 100 then we get the helong tn the pper fith pereonrililn. anawer of 78,4% Common Ideas in Reasons: It is important to note that the observation are collected randomly, hence, there is no given biased/preference for one observation over another > It's worth noting that the observations are gathered at random, so there's no prejudice or biased for one observation over another. Enduring Understanding/Generalization: Frequeny couts, grouped data, serve as pood starting of desripitive statistically analysis given a data set it gives more structured/ compact analysis and it easinr /mure convenient to use as compared to ungrouped data > When oppused to ungroaped tlata, fruuency cuurta, BrOuped data, Jerved as a uselul beginning point for deseriptive statiratically analysis given a data het. It provideu a more structured/ condensed analysin and is uusier/more
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