Question

In: Statistics and Probability

Question text In the early 1990's, the Oregon Department of Education was looking into the success...

Question text

In the early 1990's, the Oregon Department of Education was looking into the success of their school lunch programs. Critics of the current way funds were being diverted to school food services believed that the low quality of the food being served in the low-income school districts was leading to malnutrition among the students.

The state had already collected growth data for decades throughout Oregon, so they went through their records to look for signs of malnutrition. One metric they used was heights of children in the various school districts. If children did not have proper nutrition from healthy food, they would not grow to their full potential.

The CDC/National Center for Health Statistics put the average height for 12 year old girls in the United States at 59.4 inches, with standard deviation 2.3 inches.

1. Assume that heights are normally distributed. If we randomly selected a 12-year-old female student in a local school, what is the probability that she is no more than 55.4 inches tall?  

The Klamath county school district reported that their female 12-year-old students had a mean height of 58.2 inches out of a sample of 27 students. This a little more than an inch below the population mean for all 12 year old girl's heights.

2. Let's define an unusual event to be one where the probability of it occurring is less than 0.05 (or, equivalently, less than 1 in 20). If we wanted to find out if the school district's mean height was unusually low, what probability should we find?

(Hint: if 58 inches tall is unusually low, then so is 57 inches, and 56 inches.... )

Solutions

Expert Solution

1)

µ =    59.4          
σ =    2.3          
              
P( X ≤    55.4   ) = P( (X-µ)/σ ≤ (55.4-59.4) /2.3)      
=P(Z ≤   -1.74   ) =   0.0410   (answer)

2)

µ=   59.4                  
σ =    2.3                  
P(X≤x) =   0.0500                  
                      
Z value at    0.05   =   -1.6449   (excel formula =NORMSINV(   0.05   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -1.645   *   2.3   +   59.4  
X   =   55.62 (answer)          

height below 55.62 inch are unusual


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