Question

The compressive strength of samples of cement can be modeled by a normal distribution with a mean of 6000 kilograms per square centimeter and a standard deviation of 100 kilograms per square centimeter. (a) What is the probability that a samples strength is less than 6250 kilograms per square centimeter? (b) What is the probability that a samples strength is between 5800 and 5900 kilograms per square centimeter? (c) What strength is exceeded by 95% of the samples?

Answer 1

Answer

a

$\Phi (2.5) = 0.9938$

b

$P(5800 < X < 5900 ) = 0.13591$  

c

$x = 5835.5$

Question

From the question we are told that

   The mean is  $\mu = 6000 \ kg/cm^2$

    The standard deviation is  $\sigma = 100 \ kg/ cm^2$

Generally the probability that a samples strength is less than 6250 kilograms per square centimeter is mathematically represented as

        $P(X < 6250 ) = P(\frac{ X - \mu}{\sigma } < \frac{6250 - 6000 }{100} )$

Generally $\frac{X - \mu }{\sigma } = Z(The \ standardized \ value \ of \ X )$

         $P(X < 6250 ) = P(Z< 2.5 )$

From the z-table  

      $\Phi (2.5) = 0.9938$

Generally the probability that a samples strength is between 5800 and 5900 kilograms per square centimeter is mathematically represented as

    $P(5800 < X < 5900 ) = P(\frac{5800 - 6000}{100} < \frac{X -\mu}{\sigma} < \frac{5900 - 6000}{100} )$

=> $P(5800 < X < 5900 ) = P(-2< Z< -1 )$

=> $P(5800 < X < 5900 ) = P(Z < -1) - P( Z< -2 )$  

From the z-table  

    $P(Z < -2) = 0.02275$

and

    $P(Z < -1) = 0.15866$

So

    $P(5800 < X < 5900 ) = 0.15866 - 0.02275$    

=>    $P(5800 < X < 5900 ) = 0.13591$  

Generally the strength that is exceeded by 95% is mathematically evaluated  as

        $P(X > x) = 1- P(\frac{X - \mu }{\sigma} \le \frac{x- 6000}{100}) = 0.95$

=> $P(X > x) = P(Z \e \frac{x- 6000}{100}) = 0.05$

From the normal distribution table  the critical value  of  0.05 is  

    $z = - 1.645$

So

      $\frac{x- 6000}{100} = - 1.645$

=>   $x = (-1.645 * 100) + 6000$

=>   $x = 5835.5$