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. The probability of a detasseler making less than $17/hr is ≈ WHAT NUMBER %

The z-scores for a deta seller making $13 and $17 an hour will be [tex]Z_{13}=0.5 \ \ \ Z_{17}=2.5[/tex]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.[tex]Z=\dfrac{X-\mu}{\sigma}[/tex]Where x is the value of the data, μ is the mean and σ is the standard deviationIn this case :μ = 12 $/h[tex]\sigma[/tex] = 2 $/hWe need to calculate the Z-scores for x=17 and x=13[tex]Z_{13}=\dfrac{13-12}{2}=0.5[/tex][tex]Z_{17}=\dfrac{17-12}{2}=2.5[/tex]Thus the z-scores for a deta seller making $13 and $17 an hour will be [tex]Z_{13}=0.5 \ \ \ Z_{17}=2.5[/tex]To know more about z-Value follow