Respuesta :
A random number generator picks a number from 6 to 66 in a uniform manner, its resultants are mathematically given as
- X=36
- [tex]\sigma=5[/tex]
- p(x=19)=0.2167
- P(28 < x < 46) =0.3
- P(x > 48)=0.3
- P(x > 17 | x < 62) =0.75
- k=29.4
- h=15
What is the mean of this distribution?
Generally, the equation for Mean is mathematically given as
[tex]X=\frac{a+b}{2}\\\\Therefore\\\\X=\frac{6+66}{2}[/tex]
X=36
b)
Generally, the equation for standard deviation is mathematically given as
[tex]\sigma=\sqrt{\frac{b-a}{12}^2}\\\\Therefore\\\\\sigma=\sqrt{(\frac{66-6}{12})^2}\\\\\sigma=\sqrt{25}\\\\[/tex]
[tex]\sigma=5[/tex]
c)
The probability that the number will be exactly 19
[tex]p(x=19)=\frac{19-6}{66-6}[/tex]
p(x=19)=0.2167
d)
The probability that the number will be between 28 and 46 is P(28 < x < 46)
[tex]P(28 < x < 46) =\frac{46-6}{66-6} -(\frac{28-6}{66-6})[/tex]
P(28 < x < 46) =0.3
e)
The probability that the number will be larger than 48 is P(x > 48)
[tex]P(x > 48)=1-P(x\leq 48)[/tex]
Therefore
[tex]P(x > 48)=1-(\frac{48-6}{66-6})[/tex]
P(x > 48)=0.3
f)
[tex]P(x > 17 | x < 62) = P(x\leq 62-P(x leq 17))[/tex]
Therefore
[tex]P(x > 17 | x < 62) =\frac{62-6}{66-6}-(\frac{17-6}{66-6})[/tex]
P(x > 17 | x < 62) =0.75
g)
The 49th percentile
[tex]p(x\leqk=k.(1/66-6))[/tex]
Therefore
[tex]0.49=k*(1/(66-6))[/tex]
k=29.4
h)
The maximum for the lower quartile
P(xl)=h*1/60
0.25=h*1/60
h=15
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