Problem 1.NomlDiibion Supa that the TQ ofa randomly-aclecrd sident from a uniers

Question

Problem 1.NomlDiibion Supa that the TQ ofa randomly-aclecrd sident from a uniersiry ia normal, wirh a meam of 110 and aamdard desintion of 20 he 2nd is e next 255u, and so on and the quart e ma onir is the mean ofthe upper and ower bounds of the quart e hat-e the upper and ower bounds of the a A arr e ofdata set is one of our equa y sized intervals e g he 1st quarti e s te owest 25% ofthe dat 3td quartile and what is the mdpount ot ths quartle ton the given distrbution oflQ daa? :) si p se we want to ide mfv e middle % of rhe dati rhe nterval wrase midnent 1k the mean and inehuses ;0 ofthe dara what are the 14 per lower bonu da of thik tr terval it

Explanation / Answer

We are given hear the mean of the population mu=110

with sd=20

So., it's a normal distribution problem.

As we know that a quartile is 1/4 th or 25%tile and we are asked to find the lower and upper bound of the 3rd quartile. so here we need to understand the range of the 3rd quartile so., below I have depicted the ranges of each quartile

1st quartile : 0-25%

2nd quartile : 25-50%

3rd quartile : 50-75%

& 4th quartile : 75-100%

So., we are looking for 3rd and so we need to find the range of 50th percentile as lower bound and 75th percentile as upper bound.

So., we have the formula of Z as

Z=(X'-mu)/(sd) ---(A)

where we know mu=110 and sd=20

(a)From the Z table we need to find the Z value for the 0.5 (i.e. 50th percent) and we get Z=0 from the z table which by substituting we get

0=(X'-110)/20

so., X'=110 (lower bound)

Now to find the 75th percentile we need to look for the z value for 0.75 area and from the z table we get 0.6744898(or you can use this function in R to get the same(qnorm(0.75) will give ans as 0.6744898)

so substituting Z value in the equation(A) we get

0.6744898=(X'-110)/20

So., X'=123.4898 (Upper bound)

So., for 3rd quartile lower bound =110 and upper bound=123.4898

(b) now by the same way to find middle 30% i.e. 15% towards left and 15% towards right from the mean and so the area in z table for 155 towards left is (0.5-0.15=0.35) while for the right is (0.5+0.15=0.65)

So., using these functions in R (qnorm(0.35)&qnorm(0.65)) or looking at the Z table we get

Z value for lower bound=-0.3853205

Z value for upper bound=0.3853205

Substituting the same in eq(A) we get

-0.3853205=(X'-110)/20

X'=102.2936 (Lower bound)

&

0.3853205=(X'-110)/20

X'=117.7064 (Upper bound)

So., for 30% from center / mean the lower bound is 102.2936 and upper bound is 117.7064

Hope the above explaination has helped you in understanding the problem.Pls upvote if it has really helped. Good Luck!!

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