MST121 Spring Exam 2006-2007
MST121 Spring Exam 2006-2007
MST121 Spring Exam 2006-2007
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1
MULTIPLE CHOICE PART (WORTH 36 POINTS).
Please answer as many questions as possible. Each question is worth
4 points. In this part, your grade is the total of the best nine (9)
grades for answers to nine (9) questions, from amongst your answers
to the available thirteen (13) questions. From the given choices, a, b,
c & d, please do box the choice that solves the question best, and
write down the letter corresponding to your choice in the space
provided. Should you think that the correct answer is not among the
four choices, please write the word: NONE, for none of the above, in
the space provided.
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Question-1: An anti-derivative for the function, f (t ) = 32t − 1, is:
(a)16x2 (b)16x2 − 1 (c)16t2 (d) 16t2 − t
The answer is:
d
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Question-2: The distance traveled by a particle moving according to the
velocity, v(t) = t + 3, in the time interval, [1, 2], is:
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2
Question-6: The value of the derivative of the function, f (x) = ln(1+ x) −
x, at x = 0, is:
(a) −2 (b) −1 (c)0 (d) 1
1
The answer is: c
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Question-7: A pair of dice are tossed once, and the outcome numbers, on
each die, x and y, are recorded. The probability of the event, { x = 5 y }, is:
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Question-9: The sum of the first n positive odd integers: 1,3, 5,..., 2n-1,
is:
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Question-10: The mean for the geometric distribution, with probability
for success, p = 2/5, is:
(a) 5/2 (b) 5/3 (c) 5/4 (d) 1
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Question-11: On the interval [0, 3], the probability of the outcome,{x =1},
for the probability density function, f (x) = 1/3, is:
(a)1/3 (b) 0 (c) 1 (d)3
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Question-12: On the interval [0, 1], the mean for the probability density
function, f (x) = 2x, is:
(a) 1/2 (b) 3/5 (c) 2/3 (d) 1
The answer is:
c
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Question-13: On the interval [0, 1], the variance for the probability density
function, f (x) = 2x, is:
(a) Find the first derivative, f ′(x), for the function, f(x). Show that the
function, f(x), has no stationary points. Show that f (x) is an increasing
function. Does the function, f(x), have an inverse?.
f ′(x) = exp(x ) + 1 > 0, since x > 0. So, f(x), has no stationary points,
and is increasing, hence one-one, which means that, f(x), has an inverse.
(b ) Find, f (0), f ′(0), and the equation of the tangent line to the graph of
the function, f (x), at x = 0, if any.
f (0) = 0, and, f ′(0) = 2. The Tangent line equation is: y – 0 = 2(x – 0).
(c)Find the second derivative, f ′′ (x), of the function, f(x). Show that the
graph of the function, f(x), is concave up, with no inflection points.
The second derivative, f ′′ (x) = exp(x). Since, f ′′ (x) > 0. Therefore,
f(x), is concave up and has no changing sign points (no inflection points).
x
4
g (0) = 0, g ′(x) = f(x) = exp(x) + x –1, and, g ′(0) = f (0) = 0.
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(b) Find the first derivative, f ′(x), of the function, f(x), and its domain.
Find the stationary points of the function, f (x), if any.
The derivative, f ′(x) = 3(4x 3 – 2x 2 +1) > 0, on (−1, 1). There are no
stationary points of the function, f (x), since, f ′(x) = 0, has no solution.
(c) Find the absolute maximum and absolute minimum of the function,
f (x), in the interval, [−1, 1], if any.
Note that, since there are no stationary points for, f (x), then we only
consider the boundary points, and since we have, f (−1) = –4, and, f (1) =
4.
So, in [−1, 1], the absolute maximum is 4, attained at the point, x = 1,
while, the absolute minimum is – 4, attained at the point, x = – 1.
(d)Find the second derivative, f ′′ (x), of the function, f (x). Find the
inflection points of the function, f (x), if any.
The second derivative, f ′′ (x) = 3(8x2 – 4x) = 12x(2x– 1). The second
derivative changes sign at x = 0, and 1/2, the inflection points of f.
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Question–3: Consider a particle moving according to the velocity
function, v(t) = 1/(1+ t ), for t > 0, covering a distance described by the
trajectory, y(t), with y(0) = 5.
(a)Find the expression for the trajectory, y(t). Solve the equation, y(t) =
10.
The trajectory, y(t), is given by the anti-derivative v(t), with y(0) = 5.
y(t) = ln(1+t) +5. So the equation, y(t) = 10, implies t = exp(5) – 1.
(b) Find the expression for the acceleration, a(t) of the particle, and show
that the trajectory of the particle is concave down.
The acceleration of the particle , a(t) = v ′(t) = –1/(1+t)2 = –v2(t) < 0, so
the particle trajectory is concave down.
(d) Find the long term behavior of the velocity, v(t), and acceleration, a(t),
of the particle.
As t goes to infinity, the velocity of the particle converges to 0, and so
does the acceleration. In fact, despite the fact that the particle is
continuously slowing down, and decelerating (even faster), from what we
have seen in part c, the particle goes towards infinity and does not
stop…!
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Question–4: Consider the function, f (x) = a x, for x in [0, b], a, b > 0.
(a) Find f (0), f′(b), and show that, f(x), is increasing in the interval, [0,
b].
The value , f (0) = 0, f(b)= ab.
Since the derivative, f ′(x) = a > 0, on the interval [0, b], f (x) is
increasing.
(d) For a = 2, find the standard deviation, σ , for the density function.
The variance in this case,
1
σ 2 = − µ 2 + ∫ 2 x 3 dx = −4 / 9 + [ x 4 / 2]10 = 1 / 2 − 4 / 9 = 9 / 18 − 8 / 18 = 1 / 18.
0
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Question–5: Consider the below freezing monthly average temperature
list of a Canadian city:
–29, –29, –28, –32, –29, –31, –29, –31, –33, –30,–31,–32.
(a) Find the mode and the median for the monthly temperatue list.
Clearly, in this list, the mode is –29, while, the median = –30.5.
(b) Find the lower and upper quartiles, for the number list.
The upper quartile = –29, and the lower quartile = – (31 + 32)/2 = –31.5.
(d) Find the variance and the standard deviation for the number list.
The variance = (1/12)(1/9)[4(4)2 +3(5)2 + 3(2)2 + (8)2 + (1)2] = 2.
The standard deviation = Sqrt(variance) = 2 = 1.41 .
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Question - 6: A sample, of a flock of 144 Turkey birds of Farm A,
showed a mean life time of 370 days with a standard deviation of 16
days, while a sample, of 121 Turkey birds in Farm B, showed a mean life
time of 350 days, with a standard deviation of 20 days.
(a)Find a, 95% , confidence intervals, for the life time mean for Turkeys
in Farms A & B.
The Confidence interval,
for Farm A: (370–1.96*16/12, 370 + 1.96*16/12) = (367.39, 372.61),
for Farm B: (350–1.96*20/11, 360 + 1.96*20/11) = (346.44, 353.56).
(b) Use this data to find out whether there is a difference between the life
time means of the Turkey birds populations of brands A and B.
Establish the null, and alternative hypotheses.
Set mean life time of Turkeys from Farm A = µ A ,
Set mean life time of Turkeys from Farm B = µ B .
The null hypothesis: H0 : µ A = µ B .
The alternative hypothesis: H1 : µ A ≠ µ B .
(c)Find the Estimated Standard Error, ESE, for the sampling distribution.
The Estimated Standard Error, ESE, for the sampling distribution is given
by: ESE = 16 2 / 144 + 20 2 / 121 = 5.08. .
(d)Find the two sample Test Statistic, Z, and decide whether to accept or
reject the null hypothesis at the, 5%, significance level.
Denote the sample life time mean for Turkeys in Farm A = x A .
Denote the sample life time mean for Turkeys in Farm B = x B .
Z = [ x A − x B ] / ESE =[370 –350]/ESE = 20/5.08 = 3.93.
So in this case , Z > 1.96,
and hence, the null hypothesis: H0 is rejected at the, 5% , level.
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