JAMB Mathematics Past Questions & Answers - Page 311

1,551.

\(\begin{array}{c|c}
Class Interval & 3 - 5 & 6 - 8 & 9 - 11 \\ \hline Frequency & 2 & 2 & 2 \end{array}\). Find the standard deviation of the above distribution.

A.

√5

B.

√6

C.

√7

D.

√2

Correct answer is B

\(\begin{array}{c|c}Class Interval & 3 - 3 & 6 - 8 & 9 - 11 \\ x & 4 & 7 & 10 \\ f & 2 & 2 & 2 \\ f - x & 8 & 14 & 20 \\ |x - \bar{x}|^2 & 9 & 0 & 9 \\ |x - \bar{x}|^2 & 18 0 & 18 \end{array}\)

\(\bar{x}\) = \(\frac {\sum fx}{\sum f}\)

= \(\frac {8 + 14 + 20}{2 + 2 + 2}\)

= \(\frac{42}{6}\)

\(\bar{x}\) = 7

S.D = \(\sqrt\frac{\sum f(x - \bar{x})^2}{\sum f}\)

= \(\sqrt\frac{18 + 0 + 18}{6}\)

= \(\sqrt\frac{36}{6}\)

= \(\sqrt {6}\)

1,552.

\(\begin{array}{c|c} No & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline Frequency & 1 & 4 & 3 & 8 & 2 & 5 \end{array}\). From the table above, find the median and range of the data respectively.

A.

(8,5)

B.

(3, 5)

C.

(5 , 8)

D.

(5 , 3)

Correct answer is B

Median = \(\frac{\sum fx}{\sum f}\)

\(\begin{array}{c|c}
No & 0 & 1 & 2 & 3 & 4 & 5 \\ F & 1 & 4 & 3 & 8 & 2 & 5 \\ fx & 0 & 4 & 6 & 24 & 8 & 25 \end{array}\)

\(\sum fx\) = 0 + 4 + 6 + 24 + 8 + 25 = 67

\(\sum f\) = 23

Median = \(\frac{\sum fx}{\sum f}\) = \(\frac{67}{23}\) = 2.913

= \(\approx\) 3

Range = 5 - 0 = 5

(3, 5)

1,553.

The sum of four consecutive integers is 34. Find the least of these numbers

A.

7

B.

6

C.

8

D.

5

Correct answer is A

Let the numbers be a, a + 1, a + 2, a + 3

a + a + 1 + a + 2 + a + 3 = 34

4a = 34 - 6

4a = 28

a = \(\frac{28}{4}\)

= 7

The least of these numbers is a = 7

1,554.

Find \(\int\) cos4 x dx

A.

\(\frac{3}{4}\) sin 4x + k

B.

-\(\frac{1}{4}\) sin 4x + k

C.

-\(\frac{3}{4}\) sin 4x + k

D.

\(\frac{1}{4}\) sin 4x + k

Correct answer is D

\(\int\) cos4 x dx

let u = 4x

\(\frac{dy}{dx}\) = 4

dx = \(\frac{dy}{4}\)

\(\int\)cos u. \(\frac{dy}{4}\) = \(\frac{1}{4}\)\(\int\)cos u du

= \(\frac{1}{4}\) sin u + k

= \(\frac{1}{4}\) sin4x + k

1,555.

Evaluate \(\int^{1}_{0}\)(3 - 2x)dx

A.

33m

B.

5

C.

2

D.

6

Correct answer is C

\(\int^{1}_{0}\)(3 - 2x)dx

[3x - x\(^2\)]\(_{0} ^{1}\)

[3(1) - (1)\(^2\)] - [3(0) - (0)\(^2\)]

(3 - 1) - (0 - 0) = 2 - 0

= 2