Question

If the average yield of cucumber acre is 800 kg, with a variance 1600 kg, and that the amount of the cucumber follows the normal distribution. 1- What percentage of a cucumber give the crop amount between 778 and 834 kg? 2- What the probability of cucumber give the crop exceed 900 kg ?

Answer 1

Answer:

a

   The  percentage is

            $P(x_1 < X < x_2 ) = 51.1 \%$

b

   The probability is  $P(Z > 2.5 ) = 0.0062097$

Step-by-step explanation:

From the question we are told that

        The  population mean is  $\mu = 800$

        The  variance is  $var(x) = 1600 \ kg$

        The  range consider is  $x_1 = 778 \ kg \ x_2 = 834 \ kg$

         The  value consider in second question is  $x = 900 \ kg$

Generally the standard deviation is mathematically represented as

        $\sigma = \sqrt{var (x)}$

substituting value

        $\sigma = \sqrt{1600}$

       $\sigma = 40$

The percentage of a cucumber give the crop amount between 778 and 834 kg  is mathematically represented as

       $P(x_1 < X < x_2 ) = P( \frac{x_1 - \mu }{\sigma} < \frac{X - \mu }{ \sigma } < \frac{x_2 - \mu }{\sigma } )$

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

So

      $P(x_1 < X < x_2 ) = P( \frac{778 - 800 }{40} < Z< \frac{834 - 800 }{40 } )$

      $P(x_1 < X < x_2 ) = P(z_2 < 0.85) - P(z_1 < -0.55)$

From the z-table  the value for  $P(z_1 < 0.85) = 0.80234$

                                            and $P(z_1 < -0.55) = 0.29116$  

So

             $P(x_1 < X < x_2 ) = 0.80234 - 0.29116$

             $P(x_1 < X < x_2 ) = 0.51118$

The  percentage is

            $P(x_1 < X < x_2 ) = 51.1 \%$

The probability of cucumber give the crop exceed 900 kg is mathematically represented as

             $P(X > x ) = P(\frac{X - \mu }{\sigma } > \frac{x - \mu }{\sigma } )$

substituting values

             $P(X > x ) = P( \frac{X - \mu }{\sigma } >\frac{900 - 800 }{40 } )$

             $P(X > x ) = P(Z >2.5 )$

From the z-table  the value for  $P(Z > 2.5 ) = 0.0062097$