TS ICET 2019 Question Paper Shift-1 (23rd May)

The mean of first n natural numbers = [n(n+1)]/2

The mean of square of first n natural nos.=[(n)(n+1)(2n+1)]/6

                                                                          = [2n^2 + 3n + 1]/6

The mean of cube first n natural numbers = {[n(n+1)]/2}^2

The median is a simple measure of central tendency. To find the Median, we arrange the observations in order from smallest to largest value. 

If there is an odd number of observations, the median is the middle value.

If there is an even number of observations, the median is the average of two middle values.

We also need to calculate the cumulative frequency which is the sum of two frequencies.

x      fx      CF

3      5       5

5      11     5+11=16 

**7      15     16+15=31**

9      18     31+18=49

11      8      49+8=57

12      3       57+3=60

Now, (60/2) + 1 = 31

31 occurs in the 3rd column corresponding to which the number is 7.

Thus, Median = 7     Ans.

The Mode of a set of data values is the value that appears most often.

In this set of data x=2 appears most often i.e., 28 times .

Therefore, Mode of the following distribution is 2.    Ans.

Here,

Series 1 is an Arithmetic Progression with common difference = 2

Series 2 is an Arithmetic Progression with common difference = 12

Thus, Series 2 has common difference 6 times of the first.

Also, both the series have same no. of terms 

   Series 1 = (151-5)/D = (146)/2 = 73

   Series 2 = (908 - 32)/D = (876/12)= 73

Therefore , the Standard deviation of the second series will be 6 times of the first.

Hence,  6$$\sigma\ $$    Ans.

As, any five consecutive integers are required, for the sake of easy calculation take the integers as 1,2,3,4,5

Standard deviation , Sigma = $$\sqrt{\ }$$[(x - x bar ) ^ 2]/N

x     Deviation

1      (1-3)^2=4

2      (2-3)^2=1

3      (3-3)^2=0

4      (4-3)^2=1 

5      (5-3)^2=4

      where xbar = 3 (assumed mean)

Sum of X bar = 4+1+0+1+4 = 10

Therefore, $$\sigma\ $$ = $$\sqrt{\ }$$(10/5)

                   = $$\sqrt{\ }$$2    Ans.

Given, Squares of deviation, D^2= 16

No. of subjects, N=5

Now, according to Spearmen's rank correlation

Coeffiecient of rank correlation ,R = 1 - (6 *Summation D^2)/[N(N^2 -1)]

= 1 - (6*16)/[5(25-1)]

= 1 - (6*16/(5*24)

= 1 - (4/5)

=1/5   Ans.

Ocurrence of Heads/Tails appearing on a coin tossed is independent of the previous results.

Therefore , the probability of head appearing on fifth toss is same as any other time = (1/2) Ans.

Numbers out of the first 120 that are divisible by 5 = (120/5)=24

Numbers out of the first 120 that are divisible by 8 = (120/8)=15

So, the numbers that are divisible by 5 or 8 = 24+15= 39

But, these 39 numbers also includes the numbers which are divisible by both 5 and 8.

Therefore we need to subtract these numbers.

Numbers out of the first 120 that are divisible by both 5 and 8

= 120/(5*8)= 120/40=3

  Thus , Numbers out of the first 120 that are divisible by 5 or 8 = 39-3=36

Probability of the event required =36/120

                                                        =3/10  Ans.

Probability of an event, P(E)=(No. of favourable outcomes)/(No. of total outcomes)

Now, 

Total no. of outcomes  of an event with two possible results = 2^n (2 raise to power n)

                                   where n is no. of times the event occurs

Here, as the coin is tossed 5 times 

Therefore, the no. of total outcomes = 2^5 = 2*2*2*2*2 = 32

As,Total cases are 32 and we need the cases with atleast one head

So, the only case possible with no head is the case where all the coins have Tails. 

Thus, favourable outcomes are 32-1 = 31

    where 1 was the event with 5 tails.

Hence, P(E)=31/32   Ans.

Since,

P(A union B)=P(A) + P(B) - P(A intersection B)

               =(3/8) + (5/8) - (1/4)

               =1-(1/4)

               = (3/4) Ans.