During a behavioral experiment in a psychology class, each student is

Question

During a behavioral experiment in a psychology class, each student is asked to compute his or her lucky number by raising 7 to the power of the student's favorite day of the week (numbered 1 through 7 for Monday through Sunday respectively), multiplying the result by 3, and adding this to the doubled age of the student in years, rounded to the nearest year. If a class consists of 28 students, what is the probability that the median lucky number in the class will be a non-integer?

(A) 0% 
(B) 10% 
(C) 20% 
(D) 30% 
(E) 40%

BrentGMATPrepNow · Accepted Answer

We need to know 2 things to answer this question.

First, each student's lucky number will ALWAYS be an  ODD INTEGER . 
We know this because... 
lucky number = 7^(student's favorite day of the week - 1,2,3.. or 7) x 3 + ( doubled  ages of students in years) 
In other words, lucky number = (ODD INTEGER x ODD INTEGER) + EVEN INTEGER 
= ODD INTEGER + EVEN INTEGER
= ODD INTEGER

Second, when we have an even number of values (28 values), the MEDIAN equals the average (arithmetic mean) of the two middle-most integers (when all of the integers are arranged in ascending order).

Since all 28 values are guaranteed to be ODD (see point #1 above), then we know that the two middle-most integers will be ODD. 
So, the median of the 28 values = (some ODD integer + some ODD integer)/2
= (an even integer)/2
= an integer. 
In other words, the median of the 28 values is GUARANTEED to be an integer.

So, P(the median of the lucky numbers will be a non-integer) = 0%

Answer: A

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manpreetsingh86 · Answer

lucky number = 3(7^x) + 2a

where a= age of the student

let 28 students be s1,s2,s3.......s28 and assume that there lucky number follows the same order i.e. s1 has the lowest lucky number and s28 has the highest lucky number.

thus median = (s14+s15)/2

now, this median will not be an integer. if one of s14 ,s15 is even an other one is odd.

now let's see. is it possible to have the lucky number as even number.

lucky number =3(7^x) + 2a. now here x will vary from (1 to 7)

= odd (because 3(7^x) will always be odd) + even (2a will always be even) = odd
since it is not possible to have lucky number as an even quantity.
thus the probability that the median will be a non- integer = 0%

rohitmanglik · Answer

Student's age is integral value. In question, we are multiplying student's age by 2 and adding some 3* 7^x something over there.

It means Even (2* age of students) + Odd (3) * 7 ^x (where x is day of week).

We know that 7^x is always odd. (e.g. 7, 49, 243, 1701, and repeat).
=> 3* odd is again odd.

2 * age of students is always even.

So final equation is even +odd = odd.

Avg of any number of odd digits or any number of even digits can never be non integral value. So 0% is the answer.

ScottTargetTestPrep · Answer

Let’s analyze the number each students gets. When 7 raised to any positive integer power, the result will be odd. When that result is multiplied by 3, the result will be odd again. Finally, when that result is added to doubled age of the student’s age (which is an even number), the final result will always be an odd number. So each student will get an odd number regardless of his/her favorite day of the week and his/her age.

We are given that there are 28 students in the class. The median will the be the average of the the 14th and 15th numbers if the numbers are in ascending (or descending) order. However, since each number is odd and the average of two odd number is always an integer, the probability that the median number will be a non-integer is 0.

Answer: A

m7runner · Answer

→ There are 28 sample values
→ Thus there are two middle terms
→ Note that the calculation of each term is of the form: 3(7^x) + 2k, where is x = {1,.....,7}, and k is the value of the student. The first term here will ALWAYS be odd and the second term will ALWAYS be even. When added, the resultant will ALWAYS be ODD. Thus the entire sample will consist of a set of ODD samples.
→ Since we know there are 28 terms and there will be two middle terms, hence the average of these terms will be the median. Thus ODD + ODD / 2 = EVEN / 2 = EVEN, hence there will be a 0% probability of the median lucky number being a non-integer.

AnjaliSK · Answer

Hi All,

I have opted for a lil different perspective here,

Since, we have got 28 students and they can choose the values only from 1^7 to 7^7, there are only 7 values to choose from and median for odd number of values is always an integer, so that way the answer has to be 0%.

Please let me know if this way of thinking for such questions is right.

Thank you for any help in advance.

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