Answer
a
[tex]\Phi (2.5) = 0.9938[/tex]
b
[tex]P(5800 < X < 5900 ) = 0.13591 [/tex]
c
[tex]x = 5835.5 [/tex]
Question
From the question we are told that
The mean is [tex]\mu = 6000 \ kg/cm^2 [/tex]
The standard deviation is [tex]\sigma = 100 \ kg/ cm^2 [/tex]
Generally the probability that a samples strength is less than 6250 kilograms per square centimeter is mathematically represented as
[tex]P(X < 6250 ) = P(\frac{ X - \mu}{\sigma } < \frac{6250 - 6000 }{100} )[/tex]
Generally [tex]\frac{X - \mu }{\sigma } = Z(The \ standardized \ value \ of \ X )[/tex]
[tex]P(X < 6250 ) = P(Z< 2.5 )[/tex]
From the z-table
[tex]\Phi (2.5) = 0.9938[/tex]
Generally the probability that a samples strength is between 5800 and 5900 kilograms per square centimeter is mathematically represented as
[tex]P(5800 < X < 5900 ) = P(\frac{5800 - 6000}{100} < \frac{X -\mu}{\sigma} < \frac{5900 - 6000}{100} )[/tex]
=> [tex]P(5800 < X < 5900 ) = P(-2< Z< -1 )[/tex]
=> [tex]P(5800 < X < 5900 ) = P(Z < -1) - P( Z< -2 )[/tex]
From the z-table
[tex]P(Z < -2) = 0.02275[/tex]
and
[tex]P(Z < -1) = 0.15866[/tex]
So
[tex]P(5800 < X < 5900 ) = 0.15866 - 0.02275[/tex]
=> [tex]P(5800 < X < 5900 ) = 0.13591 [/tex]
Generally the strength that is exceeded by 95% is mathematically evaluated as
[tex]P(X > x) = 1- P(\frac{X - \mu }{\sigma} \le \frac{x- 6000}{100}) = 0.95 [/tex]
=> [tex]P(X > x) = P(Z \e \frac{x- 6000}{100}) = 0.05 [/tex]
From the normal distribution table the critical value of 0.05 is
[tex]z = - 1.645[/tex]
So
[tex]\frac{x- 6000}{100} = - 1.645[/tex]
=> [tex]x = (-1.645 * 100) + 6000[/tex]
=> [tex]x = 5835.5 [/tex]
