Answer:
The percentage is %z [tex]= 41.9[/tex]%
Step-by-step explanation:
From the question we are told that
The mean is [tex]\mu = 63.5 \ g[/tex]
The standard deviation is [tex]\sigma = 12.2 \ g[/tex]
The random number is x = 66 g
Given the the population is normally distributed
The probability is mathematically represented as
[tex]P(X > 66 ) = P(\frac{X - \mu }{\sigma} > \frac{x - \mu }{\sigma } )[/tex]
Generally the z-score for this population is mathematically represented as
[tex]Z = \frac{ X - \mu}{ \sigma}[/tex]
So
[tex]P(X > 66 ) = P(Z > \frac{66 - 63.5 }{12.2 } )[/tex]
[tex]P(X > 66 ) = P(Z > 0.2049 )[/tex]
Now the z-value for 0.2049 from the standardized normal distribution table is
[tex]z = 0.41883[/tex]
=> [tex]P(X > 66 ) = 0.41883[/tex]
The percentage is
% z [tex]= 0.41883 * 100[/tex]
%z [tex]= 41.9[/tex]%
