Answer:
a
The percentage is
[tex]P(x_1 < X < x_2 ) = 51.1 \%[/tex]
b
The probability is [tex]P(Z > 2.5 ) = 0.0062097[/tex]
Step-by-step explanation:
From the question we are told that
The population mean is [tex]\mu = 800[/tex]
The variance is [tex]var(x) = 1600 \ kg[/tex]
The range consider is [tex]x_1 = 778 \ kg \ x_2 = 834 \ kg[/tex]
The value consider in second question is [tex]x = 900 \ kg[/tex]
Generally the standard deviation is mathematically represented as
[tex]\sigma = \sqrt{var (x)}[/tex]
substituting value
[tex]\sigma = \sqrt{1600}[/tex]
[tex]\sigma = 40[/tex]
The percentage of a cucumber give the crop amount between 778 and 834 kg is mathematically represented as
[tex]P(x_1 < X < x_2 ) = P( \frac{x_1 - \mu }{\sigma} < \frac{X - \mu }{ \sigma } < \frac{x_2 - \mu }{\sigma } )[/tex]
Generally [tex]\frac{X - \mu }{ \sigma } = Z (standardized \ value \ of \ X)[/tex]
So
[tex]P(x_1 < X < x_2 ) = P( \frac{778 - 800 }{40} < Z< \frac{834 - 800 }{40 } )[/tex]
[tex]P(x_1 < X < x_2 ) = P(z_2 < 0.85) - P(z_1 < -0.55)[/tex]
From the z-table the value for [tex]P(z_1 < 0.85) = 0.80234[/tex]
and [tex]P(z_1 < -0.55) = 0.29116[/tex]
So
[tex]P(x_1 < X < x_2 ) = 0.80234 - 0.29116[/tex]
[tex]P(x_1 < X < x_2 ) = 0.51118[/tex]
The percentage is
[tex]P(x_1 < X < x_2 ) = 51.1 \%[/tex]
The probability of cucumber give the crop exceed 900 kg is mathematically represented as
[tex]P(X > x ) = P(\frac{X - \mu }{\sigma } > \frac{x - \mu }{\sigma } )[/tex]
substituting values
[tex]P(X > x ) = P( \frac{X - \mu }{\sigma } >\frac{900 - 800 }{40 } )[/tex]
[tex]P(X > x ) = P(Z >2.5 )[/tex]
From the z-table the value for [tex]P(Z > 2.5 ) = 0.0062097[/tex]
