Answer:
The percentage is [tex]P(0.09 < \mu < 1.2 ) = 34.9\%[/tex]
Step-by-step explanation:
From the question we are told that
The random number is [tex]x_1 = 0.09 \ and \ x_2 = 1.2[/tex]
Generally the mean of standard normal distribution is [tex]\mu = 0[/tex]
The standard deviation of a standard normal distribution is [tex]\sigma = 1[/tex]
The percentage of the data in a standard normal distribution lies between
[tex]x_1 = 0.09 \ and \ x_2 = 1.2[/tex] is mathematically represented as
[tex]P(x_1 < \mu < x_2 ) = P(\frac{x_1 - \mu}{\sigma } <\frac{X - \mu}{\sigma } < \frac{x_2 - \mu}{\sigma } )[/tex]
[tex]P(0.09 < \mu < 1.2 ) = P(\frac{0.09 - 0}{1 } <\frac{X - \mu}{\sigma } < \frac{1.2 - 0}{1 } )[/tex]
[tex]P(0.09 < \mu < 1.2 ) = P(0.09 <\frac{X - \mu}{\sigma } < 1.2 )[/tex]
The [tex]\frac{X - \mu}{\sigma } = Z(The \ standardized \ value \ of \ X )[/tex]
[tex]P(0.09 < \mu < 1.2 ) = P(0.09 <Z < 1.2 )[/tex]
[tex]P(0.09 < \mu < 1.2 ) = P (Z < 1.2 )- P( Z<0.09)[/tex]
From the z-table
[tex]P (Z < 1.2 )= 0.88493[/tex]
[tex]P( Z<0.09) = 0.53586[/tex]
So
[tex]P(0.09 < \mu < 1.2 ) = 0.88493 - 0.53586[/tex]
[tex]P(0.09 < \mu < 1.2 ) = 0.349[/tex]
Hence the percentage is
[tex]P(0.09 < \mu < 1.2 ) = 34.9\%[/tex]
