1 Answer

Urbushey · Answer 1 · 2021-08-03T15:17:32+0000

Answer:

The percentage is
$P(0.09 < \mu < 1.2 ) = 34.9\%$

Explanation:

From the question we are told that

The random number is
$x_1 = 0.09 \ and \ x_2 = 1.2$

Generally the mean of standard normal distribution is
$\mu = 0$

The standard deviation of a standard normal distribution is
$\sigma = 1$

The percentage of the data in a standard normal distribution lies between

$x_1 = 0.09 \ and \ x_2 = 1.2$ is mathematically represented as

$P(x_1 < \mu < x_2 ) = P((x_1 - \mu)/(\sigma ) <(X - \mu)/(\sigma ) < (x_2 - \mu)/(\sigma ) )$

$P(0.09 < \mu < 1.2 ) = P((0.09 - 0)/(1 ) <(X - \mu)/(\sigma ) < (1.2 - 0)/(1 ) )$

$P(0.09 < \mu < 1.2 ) = P(0.09 <(X - \mu)/(\sigma ) < 1.2 )$

The
$(X - \mu)/(\sigma ) = Z(The \ standardized \ value \ of \ X )$

$P(0.09 < \mu < 1.2 ) = P(0.09 <Z < 1.2 )$

$P(0.09 < \mu < 1.2 ) = P (Z < 1.2 )- P( Z<0.09)$

From the z-table

P (Z < 1.2 )= 0.88493

P( Z<0.09) = 0.53586

So

$P(0.09 < \mu < 1.2 ) = 0.88493 - 0.53586$

$P(0.09 < \mu < 1.2 ) = 0.349$

Hence the percentage is

$P(0.09 < \mu < 1.2 ) = 34.9\%$

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