Solved TMA questions on statistics and probability

This text provides solutions to TMA questions on statistics and probability. The questions cover topics such as sample mean, median, variance, standard deviation, cumulative distribution function, confidence interval, and more. The solutions are suitable for Arab Open University students. Desklib is an online library that offers solved assignments, essays, dissertations, and study material.

Arab Open University
Tutor Marked Assignment (TMA)
Question 1:
a) The sample mean of the charged fees is: = (Σ xi ) / n.
52+54 +55+55+55+56 +66+68+70+89
10 =620
10 =¿$62
The median of the charged fees is: {(n + 1) ÷ 2} th
= $55.5
The data set is skewed to the right because the sample mean is greater than the median of
charged fees.
b) IQR = Q3 – Q1
Where, Q1= 1
4 ( n+1 ) = 55
Q3= 3
4 ( n+ 1 ) = 67.5
 IQR= 67.5 – 55 = 12.5
Low = 55 – (1.5*12.5) = 36.25
High = 67.5 - (1.5*12.5) = 86.25
None of the given numbers can be considered an outlier because they all lie between the
limits [36.25, 86.25].
c) The sample variance is: s2= ∑ (x−x)2
n−1
s2= ( 52−62 )2 + ( 54−62 )2 + ( 55−62 )2+ (55−62 )2 + ( 55−62 )2 + ( 56−62 )2 + ( 66−62 )2+ ( 70−62 )2 + ( 89−62 )2
9
= 1192
9 = 132.44
d) The data set represented in the box plot is skewed to the right.

Question 2:
a) The expected value of no-shows is: E(X) = ∑ [ xi∗P ( x ) ]
E(X) = 0 x 0.4 + 1 x 0.3 + 2 x 0.2 + 3 x 0.1 = 0 + 0.3 + 0.4 + 0.3 = 1.0
b) The standard deviation of the no-shows is: σ ¿ √ ∑ {[xi−E ( x ) ]2∗P ¿ ¿ ¿
σ = √ (0−1.0)2 x 0.4+(1−1.0)2 x 0.3+( 2−1.0)2 x 0.2+(3−1.0)2 x 0.1
= 0.4 + 0 + 0.2 + 0.4 = 1.0
c) The cumulative distribution function is given by:
F ( x ) =
{ 0 : X<0
0.4 :0 ≤ X <1
0.7 :1 ≤ X <2
0.9 :2 ≤ X <3
1: 3≤ X
d) The probability for no more than two no-shows up is: P (X ≤ 2) = 0.7
Question 3:
Average, μ = 0.375 inches; Standard deviation, σ = 0.050 inch. To standardize, for the standard
normal distribution we use the formula z= X −μ
σ
a) P (0.36 ≤ X ≥ 0.40) = P (-0.30 ≤ Z ≥ 0.5) = 0.3094
z1 = 0.36 – 0.375
0.05 = -0.30; z2 = 0.40 – 0.375
0.05 = 0.5
=> P (Z ≥ 0.5) = 1 – P (Z ≤ 0.5) = 1 – 0.6915 = 0.3085
P (Z ≤ -0.30) = 0.6179
 P (-0.30 ≤ Z ≥ 0.5) = 0.6179 – 0.3085 = 0.3094
b) n = 50; P ( X > 0.390) is given by:
z =
0.390−0.375
( 0.05
√50 ) = 2.121
P (Z > 2.121) = 0.017

Question 4:
Sample mean = $550; sample standard deviation = $120, sample size = 20
a) 90% confidence interval for the population mean is given by:
Significance level, α = 1 – 0.9 = 0.1
Degrees of freedom, d.f. = n -1 = 20 – 1 = 19
Critical value from the t-distribution table = t0.05,19= 1.729
Margin of Error = t∗σ
√ n = 1.729 * $120/√20 = 46.39
Lower limit = $550 – 46.39 = $503.61
Upper limit = $550 + 46.39 = $596.39
 The 90% confidence interval for the population mean = [$503.61, $596.39]
b) It means that we expect that the true value of the population mean of daily revenue to fall
within $503.61 and $596.39, 90% of the time.
Question 5:
Proportion of young adults who recently purchased a home received help from their parents is:
8/40 = 0.2
a) The 95% confidence interval is given by:
Standard error = Z∗
√ ρ (1−ρ)
n = √ 0.2(1−0.2)
40 = 0.063
Critical value = Z0.05 = 1.96
 Margin of error = critical value * standard error = 1.96 * 0.063 = 0.12
Upper limit = 0.2 + 0.12 = 0.22
Lower limit = 0.2 – 0.12 = 0.08
Therefore, the 95% confidence interval for the population proportion of all young adults
who received help from their parents to get homes is [0.08, 0.22]
b) It is expected that the true population proportion of all young adults who received help
from their parents to get homes will fall within 8% and 22%, 95 percent of the time.

Question 6:
The picture below is the result obtained from the online tool http://www.physics.csbsju.edu/cgi-
bin/stats/Paired_t-test_form.sh?nrow=10.