Stat 353 Exam I Solution
SZ
2/27/2023
Q1
What is the Mean of the data: 89, 80, 92, 95, 75, 82, 90, 88, 90, 60?
Solution.
84.1
Q2
The sample standard deviation of the data: 89, 80, 92, 95, 75, 82, 90, 88, 90, 60 is ___. Round to two decimal places.
Solution.
\(s^2=\frac{\sum x^2-n(\bar{x})^2}{n-1}=\frac{71703-10(84.1)^2}{10-1}=108.32\)
\(s=10.41\)
Q3
What is the median of the sample data: 89, 80, 92, 95, 75, 82, 90, 88, 90, 60?
Solution.
Sorted data: 60 75 80 82 88 89 90 90 92 95
The median is \((88+89)/2=88.5\)
Q4
If the first quartile and third quartile of a data sample are 12 and 23, respectively, what is the IQR?
Solution.
\(23-12=11\)
Q5
If the 95th percentile of a data sample is 75, then which ofthe following is true?
Solution.
75% of the data are greater than 95.
75% of the data are less than 95.
95% of the data are less than or equal to 75.
95% of the data are greater than or equal to 75.
95% of the data are greater than 75.
(f)75% of the data are greater than or equal to 95.
Q6
Which of the following is widely used in quality and process control improvement studies where the data usually represent different types of defect, failure modes, or other categories of interest to the analyst?
Answer: pareto chart
Q7
When examining a time series plot, we look for ___ and ___.
Answer: trend, cyclic variation
Q8
Given the following data:
The sample correlation coefficient between x and y is ____ (to two decimal places).
Solution.
Using the r code:
x=1:5
y=c(2,5,9,12,18)
cor(x,y)The answer is 0.99.
Q9
Given the above probability mass function for random variable X, what is the mean?
Solution.
\(\mu=(0)(0.3)+(1)(0.1)+(2)(0.2)+(3)(0.4)=1.7\)
Q10
Given the above probability mass function for random variable X, what is the standard deviation? Round to two decimal places.
Solution.
\(\sigma^2=(0^2)(0.3)+(1^2)(0.1)+(2^2)(0.2)+(3^2)(0.4)-1.7^2=1.61\)
\(\sigma=1.27\)
Q11
Solution.
\(P(X\le 20)=1-0.5=0.5\), so
\(P(15<X\le 20)=0.5-0.3 = 0.2\)
Q12
If the probability density function of random variable is \(f(x)=2x, ~~0<x<1\), determine the standard deviation. Round to two decimal places.
Solution.
The mean is \(\mu = \int_{-\infty} ^{\infty} x\cdot 2xdx= \int_{0} ^{1} x\cdot 2xdx=\frac{2}{3}\), so \(\sigma^2=\int_{-\infty} ^{\infty} x^2\cdot 2xdx - \mu^2=\int_{0} ^{1} x^2\cdot 2xdx-(\frac{2}{3})^2=\frac{1}{2}-\frac{4}{9}=\frac{1}{18}\) and the standard deviation is 0.24.
## Q13
If the probability density function of random variable is \(f(x)=2x, ~~0<x<1\), determine \(P(X>0.5)\).
Solution.
\(P(X>0.5)=\mu = \int_{0.5} ^{1} 2xdx= 0.75\)
## Q14
Suppose that \(X\) has a standard normal distribution and \(X\) has a normal distribution with mean 100 and standard deviation 15. Determine \(P(X>120)\). Round to two decimal places.
Solution.
\(P(X>120)=P(Z>\frac{120-100}{15})=P(Z>1.33)=1-P(Z\le 1.33)=1-0.9082=0.0918\) or 0.09.
## Q15
A random variable has the probability density function \(f(x)=3e^{-3x}, x>0\). What is the standard deviation? Round the answer to two decimal places.
Solution.
Since it is the exponential distribution with \(\lambda=3\), the standard deviation is \(\frac{1}{\lambda}=\frac{1}{3}\) or 0.33.
Q16
Flip a fair coin 5 times. What is the probability that at least two heads are observed?
Solution.
Let \(X\) the number of heads out of 5 flips. Then, \(X\) has a Binomial distribution with \(n=5\) and \(p=0.5\). We want \(P(X\ge 2)\). \[P(X\ge 2)=1-P(X<2)=1-P(X=0)-P(X=1)\] \[=1-\binom{5}{0}0.5^0 (1-0.5)^{5-0}-\binom{5}{1}0.5^1 (1-0.5)^{5-1} =0.8125\]
Q17
If \(X\) has a Poisson distribution with mean 2, what is the probability that \(P(X=2)\)? Round the result to 4 decimal places.
Solution.
\(P(X=2)=\frac{\lambda^x}{x!}e^{-\lambda}=\frac{2^2}{2!}e^{-2}=0.2707\)
R code: dpois(2, lambda = 2)
Q18
What is the probability that the following system functions? Assume that each of the 3 individual component functions with probability 0.9.
Solution.
\(0.9\cdot 0.9\cdot 0.9=0.729\)
Q19
If \(E(X)=12\), then \(E(3X) =\)
Solution.
\(E(3X) =3E(X)=36\).
Q20
If \(V(X)=12\), then \(V(3X-2)=\)
Solution.
\(V(3X-2)=V(3X)=9V(X)=108\).
Q21
Given the following joint probability mass function,
Determine \(E(XY)\).
Solution.
\(E(XY)=(1)(0)(0.1)+(1)(1)(0.3)+(2)(0)(0.1)+(2)(1)(0.2)+(2)(2)(0.3)=1.9\)
Q22
Two events \(A\) and \(B\) are independent with \(P(A)=0.3\) and \(P(B)=0.6\). What is the probability that either \(A\) or \(B\) happens?
Solution.
Since \(A\) and \(B\) are independent, we have \(P(A\cap B)=P(A)P(B)=0.18\).
\(P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.6+0.3-0.18=0.72\)