Companies who design furniture for elementary school classrooms produce a variet

Question

Companies who design furniture for elementary school classrooms produce a variety of sizes for kids of different ages. Suppose the heights of kindergarten children can be described by a Normal model with a mean of 39.2 inches and standard deviation of 1.9 inches. What fraction of kindergarten kids should the company expect to be less than 35 inches tall? About of kindergarten kid are expected to be less than 35 inches tall.(Round to one decimal place as needed.) In what height interval should the company expect to find the middle 70% of kindergarteners? The middle 70% of kindergarten are expected to be between inches and inches. (Use ascending order. Round to one decimal place as needed.) At least how tall are the biggest 30% of kindergarteners are expected to be at least inches tall. (Round to one decimal place as needed.)

Explanation / Answer

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    35      
u = mean =    39.2      
          
s = standard deviation =    1.9      
          
Thus,          
          
z = (x - u) / s =    -2.210526316      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2.210526316   ) =    0.013534328 = 1.4% [ANSWER]

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b)

As the middle area is          
          
Middle Area = P(x1<x<x2) =    0.7      
          
Then the left tailed area of the left endpoint is          
          
P(x<x1) = (1-P(x1<x<x2))/2 =    0.15      
          
Thus, the z score corresponding to the left endpoint, by table/technology, is          
          
z1 =    -1.036433389      
By symmetry,          
z2 =    1.036433389      
          
As          
          
u = mean =    39.2      
s = standard deviation =    1.9      
          
Then          
          
x1 = u + z1*s =    37.23077656      
x2 = u + z2*s =    41.16922344      

Hence, between 37.2 and 41.2 [ANSWER]

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c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.7      
          
Then, using table or technology,          
          
z =    0.524400513      
          
As x = u + z * s,          
          
where          
          
u = mean =    39.2      
z = the critical z score =    0.524400513      
s = standard deviation =    1.9      
          
Then          
          
x = critical value =    40.19636097 = 40.2 in [ANSWER]      

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