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 1.9inches. a) What fraction of kindergarten kids should the company expect to be less than 39 inches tall? About % of kindergarten kids are expected to be less than 39 inches tall (Round to one decimal place as needed.) b) In what height interval should the company expect to find the middle 60% of kindergarteners? The middle 60% of kindergarteners are expected to be between » inches and inches (Use ascending order. Round to one decimal place as needed.) c) At least how tall are the biggest 25% of kindergarteners? The biggest 25% of kindergarteners are expected to be at least inches tall. (Round to one decimal place as needed.)

Explanation / Answer

here mean is 39.2 and Standard deviation is 1.9

a. P(X<39) = P(Z<39-39.2/1.9)

=P(Z<-105)

=0.4581 = 45.81% ======>>>> use excel function ''=NORMSDIST(-105)'' to get Normal probability

b.

P(X<a) = 1-0.6/2

P(X<a) = 0.2 so P(Z<-0.842) = 0.2 =====>>use NORMSINV(0.2) excel function

So here z=-0.842

a-mean/SD = -0.842

a=mean-0.842*SD

=39.2-0.842*1.9 =37.600

Now P(X<b) = 1-(1-0.6)/2

P(X<b) = 0.8 so P(Z<0.8)=0.842

So here z=0.842

a-mean/SD = 0.842

b=mean+0.842*SD

b =39.2+0.842*1.9 =40.799

The middle 60% kindergartens are expected to be in between 37.600 and 40.799.

c.

p(Z>z) = 0.25 So P(Z<z)=1-0.25=0.75

P(Z<0.674) = 0.75

z=0.674

x=mean+0.674*SD

=39.2+0.674*1.9=40.481

Hope this will be helpful. If any query please mention in cmments.

Please thumbs up if this helpful to you. God Bless You :)

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