A random number generator picks a number from 6 to 66 in a uniform manner. Round answers to 4 decimal places when possible.


The mean of this distribution is =

The standard deviation is =

The probability that the number will be exactly 19 is P(x = 19) =

The probability that the number will be between 28 and 46 is P(28 < x < 46) =

The probability that the number will be larger than 48 is P(x > 48) =

P(x > 17 | x < 62) =

Find the 49th percentile =

Find the maximum for the lower quartile =

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Answer:last one

Step-by-step explanation:

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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