2. What is common, recurring, or involved in all three problems?
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:2. What is common, recurring, or involved in all three problems?
Transcribed Image Text:B. Read the given article/texts carefully. Then, complete the table and answer the questions that follow,
Essential
Question
Text 1
Text 2
Text 3
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
movies watched in a year.
Number of
Movies
Watched
19-21
16-18
13-15
Number of
Регsons
1
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
20
13
4
3
How can
issues and
problems on
company? And at what
percent?
Answer:
your
respective
community
be solved?
Answer:
Answer:
Ore should watch at legst 1625 or 17 movies to beloe 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, (Satified) with 78,4% saifaction rate. unit.
Supporting Texts:
Supporting Texts:
Supporting Texts:
A person should wateh at least 17 movies to beleng to the
Based on the aample groupad data of 65 amployeea, the
The computer termnale in a achool 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 Sotiefied (S) on the level of sutiefaotion This les because I
we used the formula Pr bpc +In*t/100 - of /fpe 1i
The technician wanta to teet 50 urita and randomly sokecta
eome up to my awn eomputotion and formula which la
12h to be the first unit to be tasted, The second or the
where Ibpen first alass boundary in the loated cumulative weighted means VS(25) + K(20) + NC13) + SD14) +
frequeney, n value we need to find, fa frequeney of the
value (total number of persons), and ef= oumulative
next computer unit will be tested is the 18th unit This is
VD1) / total number of employees where VS= very
because we use my buit up 2 formala which is umbere
atiafind, S- autitafied, N- netral, SD- aomwhat
of planned to test computer unita total computer termhale
dissatisfied, VD= very dissatisfled. the mean is 392 or
appreximately 4 which indicate as Saliefied (S) en the
level of sutisfuction, ond total number of employeas 65,
frequency of the class before the class with the value of
n schools" and "first unit to be tested + anawer for the
the given perentie.
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
bourdariea ta oomplete the table, now we solve for the
upper Sth percentile, Upper Sth percentile is 100-5-95th
mean by getting the sum of multiplied frequency to ita'
aubstitute the values needed, There are 300 oomputer
percentile. So we reed to colesiate the 95th percentile. We numerical value and divided it by the total frequency form
use the formula "Pr Ibpo +Entf/100 - ef fpa 11", We
first find the n*1/100 95*5o cso is from the total number
terminale in a school Fifty units are reeded to test if the
the formula "weighted mean= VS(25) + S(20) + N(13) +
computers are in good working condition. Since the first unit
SD(4) + VD(1) / total number of employees". We
of persons)a 4,750/100- 47.5. Then we loeate the
to be tested is 12th, we need to divide firet the number of
eumulotive frequeney value grester or equol to 47.5. After substitute the given vales to the formula, weighted mean
dutermining, wa find the n=95, lbde-15.5, ef-47, fpe-2
eomputer terminals which is 300 to the numbers of units
needed to test which is 50, It is equal to 6 as the interval
of testing, ater that we use the first unit to be tested
,
5(25) + 4(20) + 3(13) + 2(4) + 1(1) / 65 then we get the
and i=18.5-15.5-3. We subetitute the vahes in the formula
answer of 3.92 and if we round it off, we'l get the answer
- of 4 that shows the employees' level of satiefoction of
then inturprut, P95- 15.5 (47,5- 47 / 2)3, and the
answer is 16.25. So vakue is 16.25 is in the upper 5th
percentile or the 95th parcenntile round 16.25 up te the
nearest whole number is 17 Tie io because we oon't have percent, we divided the mean to the highest value of level
which is 12 to be added to the interval of testing and the
Satisfind, And to determine the level of satiefaction in
sum is 18 which will be the next computer unit to be tested
decimcal valum tar number of movina axt is a diernin
vartable). So we must have at least 17 movies to watch to of satisfaction and multiply it by 100 then we get the
halang tn the pypar fith peronliln,
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:
Frequery couts, grouped data, serve as pood starting of desriptive
statistically analysis given a data set it gives more structured/ compact
nalysis and it easinr /mure convenient to use as compared to ungrouped
data
> When oppused to ungroaped tlata, fruuency cuurta, BrOuperd data, served
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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