Take home exercise 3. A psychologist administered a personalityinventory test for passive-aggressive traits to 150 employees.Each person was given a score from 1 to 5, where 1 is extremelypassive and 5 is extremely aggressive. The results are shown inthe table. Construct a probability distribution for the randomvariable x. Then graph the distribution using a histogram.Score, xFrequency, fWhat is the probability ofrandomly selecting an employeewith a score of 4 or 5? Between2 and 4? (inclusive)124334230521OF CEescF10#$4&2367OdeteteQWERYUPtabADFGKcaps lockreturnXCVBshiftshiftfncontroloptioncommandcommandoptionL0RksDFFOfPFAVukudLOAL FLORIDASISAIN

Question
Asked Oct 15, 2019
Take home exercise 3. A psychologist administered a personality
inventory test for passive-aggressive traits to 150 employees.
Each person was given a score from 1 to 5, where 1 is extremely
passive and 5 is extremely aggressive. The results are shown in
the table. Construct a probability distribution for the random
variable x. Then graph the distribution using a histogram.
Score, x
Frequency, f
What is the probability of
randomly selecting an employee
with a score of 4 or 5? Between
2 and 4? (inclusive)
1
24
33
42
30
5
21
OF CE
esc
F10
#
$
4
&
2
3
6
7
O
detete
Q
W
E
R
Y
U
P
tab
A
D
F
G
K
caps lock
return
X
C
V
B
shift
shift
fn
control
option
command
command
option
L0Rks
DFF
OfP
FA
Vukud
LO
AL FLORIDAS
ISAIN
help_outline

Image Transcriptionclose

Take home exercise 3. A psychologist administered a personality inventory test for passive-aggressive traits to 150 employees. Each person was given a score from 1 to 5, where 1 is extremely passive and 5 is extremely aggressive. The results are shown in the table. Construct a probability distribution for the random variable x. Then graph the distribution using a histogram. Score, x Frequency, f What is the probability of randomly selecting an employee with a score of 4 or 5? Between 2 and 4? (inclusive) 1 24 33 42 30 5 21 OF CE esc F10 # $ 4 & 2 3 6 7 O detete Q W E R Y U P tab A D F G K caps lock return X C V B shift shift fn control option command command option L0Rks DFF OfP FA Vukud LO AL FLORIDAS ISAIN

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check_circleExpert Solution
Step 1

The probability distribution for the random variable x is obtained below:

The required formula for the probability distribution for the random variable x is,

f
Σί
P(x)
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Image Transcriptionclose

f Σί P(x)

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Step 2

The probability distribution for the random variable x is tabulated below:

Score (x) Frequency (y)
P(X)
24
= 0.16
150
1
24
33
0.22
2
33
150
42
0.28
3
42
150
30
= 0.20
150
4
30
21
0.14
150
5
21
Σ1=150
Total
help_outline

Image Transcriptionclose

Score (x) Frequency (y) P(X) 24 = 0.16 150 1 24 33 0.22 2 33 150 42 0.28 3 42 150 30 = 0.20 150 4 30 21 0.14 150 5 21 Σ1=150 Total

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Step 3

The procedure to draw a Histogram in MINITAB is given below,

  1. Choose Graph > Bar Chart.
  2. From Bars represent, choose a function of a variable.
  3. Under One Y, choose Simple. Click OK.
  4. Under function choose mean.
  5. In Graph variables, enter the column of P(X).
  6. In categorical ...
Chart of P(X)
0.30
0.25
0.20
0.15
0.10
0.05
0.00
2
3
5
Score (x)
(x)d
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Image Transcriptionclose

Chart of P(X) 0.30 0.25 0.20 0.15 0.10 0.05 0.00 2 3 5 Score (x) (x)d

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