Statistics Assignment: Probability, Distributions, and Data Analysis

This statistics assignment solution covers a range of statistical concepts. It begins with a frequency distribution table, calculating missing values and cumulative relative frequency. The solution then calculates the sample variance and standard deviation from a given frequency table, using gro...Read More

STATISTICS
STUDENT NAME/ID
[Pick the date]

Question 2
Frequency distribution table
Number of seniors = 150 =
Score Range Frequency Relative Frequency Cumulative Relative
Frequency
Very anxious =0.20*150
=30
0.20 =0.20
Anxious =0.1*150 =15 =0.30-0.20=0.1 0.30
Mildly anxious =0.8*150 =12 =0.38-0.30=0.08 =0.68-0.30=0.38
Generally relaxed 45 =45/150 =0.30 =1-0.32=0.68
Very relaxed =0.32*150
=48
0.32 =1
Total 150 1
Question 3
Given frequency distribution
1

Mean ( μ ) =∑ F . M
∑ F =( 125
100 )=1.25
Now,
Variance σ2 = 1
∑ F−1 (∑ F . M2 )− (∑ F∗μ2 )
Variance σ2 = 1
99 ( 1740− ( 100∗( 1.25 ) 2 ) ) =15.997
Standard deviation σ= √15.997
Standard deviation σ=3.9997 4
Question 4
Percentage of students female =55%
Percentage of students receives grade of C =40%
Percentage of students female and not C student =35%
Contingency table
2

Probability (male, C student) =0.20/0.40 = 0.50
There is a 0.50 probability that a randomly selected student is a male and is a “C” student.
Question 5
P ( A )=0.6
P ( B ) =0.5
P ( A ∪ B ) =?
Now,
For independent events
P ( A ∪B ) =P ( A ) + P ( B ) −P ( A )∗P ( B )
P ( A ∪B ) =0.6+0.5− ( 0.6∗0.5 ) =0.8
Question 6
Number of items =15
Defective pieces = 6
When two items are randomly taken from the lot then the probability that there would be exactly
one non-defective item =?
Here, 9 items out of 15 items are non-defective and hence,
N=15 , K =9 , N =2
3

P ( X=1 )=
9
1
6
1
15
2
= 9∗6
105 =18
35 =0.5413
Therefore,
Requisite probability =0.5413
Question 7
Probability of a defective calculator = 20%
Sample size = 15
Probability that less than 5 of calculator will be defective =?
Binomial distribution
P( x <5)=P( X ≤ 4 )=0.8358(Standard binomial table)
Question 8
Likelihood of troubles = 0.75
Trouble reported = 5
Probability that at least 3 troubles would be repaired on same day =?
Binomial distribution
P( x ≥3)=1−P( x ≤ 2)=1−0.103516=0.8965(Standard binomialtable)
Question 9
Mean μ = 60 feet
Standard deviation σ =4 feet
Probability that length of athlete throws the hammer would fall between 56 feet and 65 feet =?
P(56<X<65)
4

P ( 56< X <65 )=P ( 56−60< X−μ< 65−60 )
P ( 56< X <65 )=P ( 56−60
4 < X −μ
σ < 65−60
4 )
P ( 56< X <65 )=P (−1<Z <1.25 ) =0.7357(Standard normal table)
Probability that length of athlete throws the hammer would fall between 56 feet and 65 feet is
0.7357.
Question 10
Binomial distribution
n=100
p=0.2
Probability that x is less than or equal to 15 using normal approximation to binomial distribution.
 Without continuity corrections
Mean=np=100∗0.2=20
Standard deviation ¿ √100∗0.2∗0.8=¿4
Now,
The z value= 15−20
4 =−1.25
P ( x≤15 ) =P ( z ≤−1.25 ) =0.1056 (Normal distribution table)
5