Answer :
Answer:
The number of standard deviation above the mean is [tex] z_o = 1.4607 [/tex]
Step-by-step explanation:
From the question we are told that
The mean weight of ears of corn from each farm is [tex]\mu=1.26[/tex]
The standard deviation of ears of corn from Iowa is [tex]\sigma_1[/tex]
The standard deviation of ears of corn from Ohio is
[tex] \sigma_2= 0.01 + \sigma_1 [/tex]
The weight of a random selected ear of corn from Iowa is x = 1.39 pound
The standardized score is z =1.645
The weight of a random selected ear of corn from Ohio is [tex]x_1 = 1.39\ pound[/tex]
Generally the standardized score of weight of corn from Iowa is mathematically represented as
[tex]z = \frac{ 1.39 - 1.26 }{\sigma_1 }[/tex]
=> [tex] 1.645 = \frac{ 1.39 - 1.26 }{\sigma_1 }[/tex]
=> [tex] \sigma_1 = \frac{ 1.39 - 1.26 }{ 1.64 5}[/tex]
=> [tex] \sigma_1 = 0.0790 [/tex]
Generally the standardized score of weight of corn from Ohio is mathematically represented as
[tex]z_o = \frac{ 1.39 - 1.26 }{\sigma_2 }[/tex]
=> [tex] z_o= \frac{ 1.39 - 1.26 }{0.01 + \sigma_1 }[/tex]
=> [tex] z_o = \frac{ 1.39 - 1.26 }{ 0.089}[/tex]
=> [tex] z_o = 1.4607 [/tex]
Because the value is positive it means that this value represent the number of standard deviation above the mean.
Using the normal distribution, it is found that the weight with respect to the Ohio distribution is 1.46 standard deviations from the mean.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
For the Iowa farm:
- The mean is of 1.26 pounds, hence [tex]\mu = 1.26[/tex].
- The weight is of 1.39 pounds, hence [tex]X = 1.39[/tex]
- The standardized score is of 1.645, hence [tex]Z = 1.645[/tex].
Then, the standard deviation can be found as follows:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{1.39 - 1.26}{\sigma}[/tex]
[tex]1.645\sigma = 0.13[/tex]
[tex]\sigma = \frac{0.13}{1.645}[/tex]
[tex]\sigma = 0.079[/tex]
In Ohio:
- Same mean.
- Standard deviation 0.01 pound greater, hence [tex]\sigma = 0.079 + 0.01 = 0.089[/tex]
- Same weight.
Hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1.39 - 1.26}{0.089}[/tex]
[tex]Z = 1.46[/tex]
The weight with respect to the Ohio distribution is 1.46 standard deviations from the mean.
To learn more about the normal distribution, you can check https://brainly.com/question/24663213