Question

To control pollination, pollen-producing flowers are often removed from the top of corn in a process called detasseling. The hourly rates for detasselers in Iowa are roughly normally distributed, with a mean of $12/hr and a standard deviation of $2/hr. What are the z-scores for a detasseler making $13 and $17 an hour? z13 = 0. 9, z17 = 1. 25 z13 = 0. 5, z17 = 1. 25 z13 = 0. 5, z17 = 2. 5.

Answer 1

The z-scores for a deta seller making $13 and $17 an hour will be $Z_{13}=0.5 \ \ \ Z_{17}=2.5$

What is z- value ?

The z-scores give us information about how many standard deviations from the mean the data are. This difference can be negative, if the data are n deviations to the left of the mean, or it can be positive if the data are n deviations to the right of the mean.

To calculate the Z scores, we calculate the difference between the value of the data and the mean and then divide this difference by the standard deviation.

$Z=\dfrac{X-\mu}{\sigma}$

Where x is the value of the data, μ is the mean and σ is the standard deviation

In this case :

μ = 12 $/h

$\sigma$ = 2 $/h

We need to calculate the Z-scores for x=17 and x=13

$Z_{13}=\dfrac{13-12}{2}=0.5$

$Z_{17}=\dfrac{17-12}{2}=2.5$

Thus the z-scores for a deta seller making $13 and $17 an hour will be $Z_{13}=0.5 \ \ \ Z_{17}=2.5$

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