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2 months ago

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A grocery store receives deliveries of corn from two farms, one in Iowa and the other in Ohio. Both farms produce ears of corn w

ith mean weight 1.26 pounds. The standard deviation of the weights of the ears of corn from the farm in Ohio is 0.01 pound greater than that from the farm in Iowa. A randomly selected ear of corn from the farm in Iowa weighed 1.39 pounds, which has a standardized score of 1.645 for the distribution of weights for the Iowa corn. If an ear of corn from the farm in Ohio weighs 1.39 pounds, how many standard deviations from the mean is the weight with respect to the Ohio distribution

1 answer:

Zina [12.3K]2 months ago

4 0

Answer:

The number of standard deviations above the mean is $z_o = 1.4607$

Step-by-step explanation:

The question indicates that:

The average weight of the corn ears from each farm is $\mu=1.26$

The standard deviation for the corn ears from Iowa is $\sigma_1$

The standard deviation for the corn ears from Ohio is

$\sigma_2= 0.01 + \sigma_1$

A randomly chosen ear of corn from Iowa weighs x = 1.39 pounds

The standardized score is z = 1.645

The weight of a randomly chosen ear of corn from Ohio measures $x_1 = 1.39\ pound$

In general, the standardized score of corn weight from Iowa can be mathematically defined as:

$z = \frac{ 1.39 - 1.26 }{\sigma_1 }$

=> $1.645 = \frac{ 1.39 - 1.26 }{\sigma_1 }$

=> $\sigma_1 = \frac{ 1.39 - 1.26 }{ 1.64 5}$

=> $\sigma_1 = 0.0790$

Conversely, the standardized score of corn weight from Ohio is expressed as:

$z_o = \frac{ 1.39 - 1.26 }{\sigma_2 }$

=> $z_o= \frac{ 1.39 - 1.26 }{0.01 + \sigma_1 }$

=> $z_o = \frac{ 1.39 - 1.26 }{ 0.089}$

=> $z_o = 1.4607$

A positive value indicates this quantity represents the number of standard deviations above the mean.

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