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Mathematics
UTME / JAMB 1988
Statistics and Probability
Solution
\(\begin{array}{c|c} Scores(x) & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline Frequency(f) & 7 & 11 & 6 & 7 & 7 & 5 & 3\end{array}\)
In the distribution above, the mode and median respectively are
A
1,3
B
1, 2
C
3, 3
D
0, 2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
From the distribution, Mode = 1 and
Median = \(\frac{2 + 2}{2}\)
= 2
= 1, 2
Source: Myschool Past Questions Archive (Record #31647)
Mathematics
UTME / JAMB 1988
Statistics and Probability
Solution
If x and y represent the mean and the median respectively of the following set of numbers 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, find the \(\frac{x}{y}\) correct to one decimal place
A
18
B
15
C
1.0
D
5
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Mean \(\bar{x}\) = \(\frac{156}{10}\) = 15.6
Median = \(\bar{y}\) = \(\frac{15 + 16}{2}\)
\(\frac{31}{2}\) = 15.5
\(\frac{x}{y}\) = \(\frac{15.6}{15.5}\) = 1.0065
1.0(1 d.p)
Source: Myschool Past Questions Archive (Record #31770)
Mathematics
UTME / JAMB 1988
Statistics and Probability
Solution
If two dice are thrown together, what is the probability of obtaining at least a score of 10?
A
\(\frac{1}{6}\)
B
\(\frac{1}{12}\)
C
\(\frac{5}{6}\)
D
\(\frac{11}{12}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
The total sample space when two dice are thrown together is 6 x 6 = 36
\(\begin{array}{c|c} & 1 & 2 & 3 & 4 & 5 & 6\\ 1. & 1.1 & 1.2 & 1.3 & 1.4 & 1.5 & 1.6\\ 2 & 2.1 & 2.2 & 2.3 & 2.4 & 2.5 & 2.6\\3 & 3.1 & 3.2 & 3.3 & 3.4 & 3.5 & 3.6\\ 4 & 4.1 & 4.2 & 4.3 & 4.4 & 4.5 & 4.6\\ 5 & 5.1 & 5.2 & 5.3 & 5.4 & 5.5 & 5.6\\6 & 6.1 & 6.2 & 6.3 & 6.4 & 6.5 & 6.6 \end{array}\)
At least 10 means 10 and above
P(at least 10) = \(\frac{6}{36}\)
= \(\frac{1}{6}\)
Source: Myschool Past Questions Archive (Record #31783)
Mathematics
UTME / JAMB 1987
Statistics and Probability
Solution
The goals scored by 40 football teams from three league divisions are recorded below
No of goals
0
1
2
3
4
5
6
Frequency
4
3
15
16
1
0
1
What is the total number of goals scored by all the teams?
A
21
B
40
C
91
D
96
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Let x rep. number of goals
x
f
fx
0
4
0
1
3
3
2
15
30
3
16
48
4
1
4
5
0
0
6
1
6
\(\sum fx\) = 91
Total number of goals scored is \(\sum fx\)= 91
Source: Myschool Past Questions Archive (Record #31990)
Mathematics
UTME / JAMB 1987
Statistics and Probability
Solution
The numbers 3, 2, 8, 5, 7, 12, 9, and 14 are the marks scored by a group of students in a class test. If P is the mean and Q the median, the P + Q is
A
18
B
17\(\frac{1}{2}\)
C
16
D
15
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Re-arranging them 2, 3, 5, 7, 8, 9, 12, 14
Mean = P and median = Q
P = \(\frac{2 + 2 + 5 + 7 + 8 + 9 + 12 + 14}{8}\)
\(\frac{60}{8}\) = 7.5
median (Q) = \(\frac{7 + 8}{2}\)
\(\frac{15}{2}\) = 7.5
P + Q = 7.5 + 7.5 = 15
Source: Myschool Past Questions Archive (Record #31991)
Mathematics
UTME / JAMB 1986
Statistics and Probability
Solution
Find the median of the numbers 89, 141, 130, 161, 120, 131, 131, 100, 108 and 119
A
131
B
125
C
123
D
120
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Arrange in ascending order
89, 100, 108, 119, |120, 130|, 131, 131, 141, 161
Median = \(\frac{120 + 130}{2}\)
= 125
Source: Myschool Past Questions Archive (Record #32174)
Mathematics
UTME / JAMB 1986
Statistics and Probability
Solution
A man kept 6 black, 5 brown and 7 purple shirts in a drawer. What is the probability of his picking a purple shirt with his eyes closed?
A
\(\frac{1}{7}\)
B
\(\frac{11}{18}\)
C
\(\frac{7}{18}\)
D
\(\frac{7}{11}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Prob. of purple shirt is \(\frac{7}{18}\)
Source: Myschool Past Questions Archive (Record #32177)
Mathematics
UTME / JAMB 1986
Statistics and Probability
Solution
The table below gives the scores of a group of students in a Mathematical test.
\(\begin{array}{c|c} Scores & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline Frequency & 2 & 4 & 7 & 14 & 12 & 6 & 4 & 1\end{array}\)
If the mode in m and the number of students who scored 4 or less is s. What is (s, m)?
A
(27, 4)
B
(14,4)
C
(13,4)
D
(4,4)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
M = mode = the number having the highest frequency = 4
S = Number of students with 4 or less marks
= 14 + 7 + 4 + 2
= 27
∴ (M,S) = (27, 4)
Source: Myschool Past Questions Archive (Record #32178)
Mathematics
UTME / JAMB 1985
Statistics and Probability
Solution
The number of goals scored by a football team in 20 matches is shown below
\(\begin{array}{c|c} \text{No. of goals} & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \text{No. of matches} & 3 & 5 & 7 & 4 & 1 & 0 \end{array}\)
What are the values of the mean and the mode respectively?
A
(1.75, 5)
B
(1.75, 2)
C
(1.75, 1)
D
(65, 74)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\begin{array}{c|c}x & f & fx\\ \hline 0 & 3 & 0\\ 1 & 5 & 5\\ 2 & 7 & 14\\ 3 & 4 & 12\\4 & 1 & 4\\5 & 0 & 0\end{array}\)
\(\sum{f}\) = 10
\(\sum fx\) = 35
Mean = \(\frac{\sum fx}{\sum f}\)
= \(\frac{35}{20}\)
= \(\frac{7}{4}\)
= 1.75
Mode = 2
= 1.75, 2
Source: Myschool Past Questions Archive (Record #32252)
Mathematics
UTME / JAMB 1985
Statistics and Probability
Solution
If two fair coins are tossed, what is the probability of getting at least one head?
A
\(\frac{1}{4}\)
B
\(\frac{1}{2}\)
C
1
D
\(\frac{43}{78}\)
E
\(\frac{3}{4}\)
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
Prob. of getting at least one head
Prob. of getting one head + prob. of getting 2 heads
= \(\frac{1}{4}\) + \(\frac{2}{4}\)
= \(\frac{3}{4}\)
Source: Myschool Past Questions Archive (Record #32258)
Mathematics
UTME / JAMB 1985
Statistics and Probability
Solution
A bag contains 4 white balls and 6 red balls. Two balls are taken from the bag without replacement. What is the probability that they are both red?
A
\(\frac{2}{3}\)
B
\(\frac{2}{15}\)
C
\(\frac{1}{2}\)
D
\(\frac{1}{3}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
P(R1) = \(\frac{6}{10}\)
= \(\frac{2}{3}\)
P(R1 n R11) = P(both red)
\(\frac{3}{5}\) x \(\frac{5}{9}\)
= \(\frac{1}{3}\)
Source: Myschool Past Questions Archive (Record #32278)
Mathematics
UTME / JAMB 1984
Statistics and Probability
Solution
Thirty boys and x girls sat for a test. The mean of the boys' scores and that of the girls were respectively 6 and 8. Find x if the total scores was 468
A
38
B
24
C
36
D
22
E
41
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Number of boy's = 30
Number of girls = x
Mean of boys' score = 6, for girls = 8
Total score for boys = 30 x 6
Total score for both boys and girls = 468
180 + 8x = 468
x = \(\frac{288}{8}\)
= 36
Source: Myschool Past Questions Archive (Record #32336)
Mathematics
UTME / JAMB 1984
Statistics and Probability
Solution
Two fair dice are rolled. What is the probability that both show up the same number of points.
A
\(\frac{1}{36}\)
B
\(\frac{7}{36}\)
C
\(\frac{1}{2}\)
D
\(\frac{1}{3}\)
E
\(\frac{1}{6}\)
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
A dice has 6 faces, 2 dice has 6 x 6 = 36 combined face
Prob. of both showing the same number of points
= \(\frac{6}{36}\)
= \(\frac{1}{6}\)
Source: Myschool Past Questions Archive (Record #32439)
Mathematics
UTME / JAMB 1984
Statistics and Probability
Solution
Measurements of the diameters, in centimeters, in centimeters, of 20 copper spheres are distributed as shown below
\(\begin{array}{c|c} \text{Class boundary in cm} & \text{frequency} & \\\hline 3.35 - 3.45 & 3\\ 3.45 - 3.55 & 6\\ 3.55 - 3.65 & 7\\ 3.65 - 3.75 & 4\end{array}\)
What is the mean diameter of the copper spheres?
A
3.40cm
B
3.58cm
C
3.56cm
D
3.62cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\begin{array}{c|c} \text{x(mid point)} & f & fx\\ \hline 3.4 & 3 & 10.2 \\ 3.5 & 6 & 21.0\\3.6 & 7 & 25.2\\3.7 & 4 & 14.8\end{array}\)
\(\sum f\) = 20
\(\sum fx\) = 71.2
mean = \(\frac{\sum fx}{\sum f}\)
= \(\frac{71.2}{20}\)
= 3.56
Source: Myschool Past Questions Archive (Record #32442)
Mathematics
UTME / JAMB 1983
Statistics and Probability
Solution
If M represents the median and D the mode of the measurements 5, 9, 3, 5, 7, 5, 8, then (M, D) is
A
(6, 5)
B
(5, 8)
C
(5, 7)
D
(5, 5)
E
(7, 5)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
First re-arrange the given data in the form 3, 5, 5, 5, 7, 8, 9.
3, 5, 5,|5| 7, 8, 9
median(M) = 5
mode(D) = 5
(M, D) = (5, 5)
Source: Myschool Past Questions Archive (Record #32443)