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BASKIN SCHOOL OF ENGINEERING
Department of Applied Mathematics
and Statistics
AMS 7 Spring 2015
Review Problems for Midterm Exam
Section 6.3 #9:
Find the Margin of Error and Confidence Interval for estimating the population mean µ.
Population Parameter: µ = Mean salaries of Statistic’s Professors: 95% Confidence; n =
100; X̄ = $95, 000 and σ is known to be $12, 345.
(1) Normality conditions fulfilled (n > 30).
(2) Critical Value: 1 − α = 0.95, α = 0.05; α/2 = 0.025; Area to the left of the critical
value: 1 − 0.025 = 0.975. From Table A-2, zα/2 = 1.96.
(3) Margin of Error:
σ
12, 345 1.96 × 12, 345
E = zα/2 × √ = 1.96. √
= 2, 419.62
=
10
n
100
(4) C.I.:
X̄ − E < µ < X̄ + E
92, 580.38 < µ < 97, 417.62
or
X̄ ± E
or
95, 000 ± 2, 419.62
(X̄ − E, X̄ + E)
(92580.38, 97419.62)
(5) Interpreting Results: If we were to select many different Statistic’s Professor samples
of size 100 and construct 95% confidence intervals for their mean salaries, 95% of
them would actually contain the value of µ.
Section 5.3, # 13:
a) Estimate the probability of getting more than 55 girls in 100 births. Assume that
boys and girls are equally likely.
b) It is unusual to get more than 55 girls in 100 births?
Part a)
(1) Random Variable: X = # of girls in 100 births. X is a Binomial Random Variable.
Normal Approximation to Binomial?.
Assume p = 0.5; n = 100; q = 0.5. n.p = n.q = 50 ≥ 5. (It is Ok to use the
Normal approximation)
(2) Find µ and σ for the Normal distribution.
For a Binomial Random Variable:
µ = n.p = 50
√
√
σ = n.p.q = 100 × 0.5 × 0.5 = 5
(3) Draw Normal curve. Use continuity correction.
(4) Find z Score and probability of shaded area:
z=
55.5 − 50
5.1
=
= 1.1
5
5
P (z ≥ 1.1) = 1 − 0.8643 = 0.1357
Part b)
Check if P(More than 55 girls) is greater or less than 0.05.
P (X > 55) = P (z ≥ 1.1) = 1 − 0.8643 = 0.1357 > 0.05
It is not unusual to get more than 55 girls in 100 births.
Section 5.5 #11: For women aged 18 − 24 Systolic Blood Pressure (mmHg) is normally
distributed with a mean of 114.8 and a standard deviation of 13.1. (Hypertension: BP >
140)
a) If a woman between the ages 18 − 24 is randomly selected, find the probability that
her systolic BP is greater than 140.
b) If 4 women in that age range are selected, find their probability that their mean
systolic BP is greater than 140.
Part a) Random Variable: X = Systolic Blood Pressure in women with age range of 18 − 24.
Find z Score: µ = 114.8, σ = 13.1
z=
Part b)
x−µ
140 − 114.8
25.2
=
=
= 1.9236
σ
13.1
13.1
P (Z > 1.92) = 1 − 0.9726 = 0.0274
(1) Check conditions for Central Limit Theorem: Population is normally distributed even
though n < 30 (n = 4)
(2) Find µx̄ and σx̄ . X̄ has a Normal distribution with mean µx̄ = µ and Standard
Deviation σx̄ = √σn .
µx̄ = 114.8
13.1
13.1
σx̄ = √ =
= 6.55
2
4
(3) Find z Score:
140 − 114.8
25.2
=
= 3.8473
6.55
6.55
P (z > 3.8473) = 1 − 0.9999 ≈ 0.0001
There is a 0.0001 probability that a group of 4 women will have a mean systolic
BP greater than 140.
z=
Section 4.5 #6: 11 Babies are born in Westport Village each year (Pop:760)
a) Find the mean number of births per day.
b) Find the probability that on a given day, there are no births.
c) Find the probability of at least one birth on any given day.
Random Variable: X = # of babies born in Westport Village per day.
√
X has a Poisson distribution with mean µ and Standard Deviation µ.
Part a)
µ=
11
= 0.03 babies/day
365
Part b)
µx . exp−µ
X = 0, 1, 2, . . .
P (X = x) =
x!
P (No births on a given day) = P (X = 0)
µ0 . exp−µ
=
0!
= exp−µ
= exp−0.03
= 0.9703
Part c)
P (At least one birth on a given day) = 1 − P (No births on a given day)
= 1 − 0.9703
= 0.0296
Section 3.5 #13: A surgeon wants to assure he is not late for an early operation because
of malfunction of her alarm clock. She decides to use three clocks. What is the probability
that at least one of her alarm clocks works correctly if each individual alarm clock has a 99%
chance of working correctly?
Events: C1 = Clock 1 works fine; C2 = Clock 2 works fine; C3 = Clock 3 works fine. Events
are independent.
P (C1) = 0.99 → P (C̄1) = 1 − P (C1 ) = 0.01
(Prob. Clock 1 does not work fine)
P (C2) = 0.99 → P (C̄2) = 1 − P (C2 ) = 0.01
(Prob. Clock 2 does not work fine)
P (C3) = 0.99 → P (C̄3) = 1 − P (C3 ) = 0.01
(Prob. Clock 3 does not work fine)
P (One of her alarm clocks works correctly) =
=
=
=
=
1 − P (None of her alarm clocks work correctly)
1 − P (C̄1 &C̄2 &C̄3 )
1 − P (C̄1 ) × P (C̄2) × P (C̄3 )(Using Mult. Rule)
1 − (0.01)3
0.9999
The doctor would be better of by using 3 clocks!!!!