BUS2030 Statistics Assignment: Distributions and Confidence Intervals

This document provides a complete solution to a Statistics for Business (BUS2030) assignment. The solution covers a range of statistical concepts including the normal distribution, uniform distribution, sampling distribution of the mean, and confidence intervals. Question 1 focuses on the normal distribution, calculating probabilities and identifying values within a specified range. Question 2 also deals with the normal distribution, determining values corresponding to specific percentages. Question 3 explores the uniform distribution, calculating probabilities for time intervals and determining mean and standard deviation. Question 4 addresses the sampling distribution of the mean, calculating probabilities related to sample means. Question 5 and 6 delve into confidence interval estimation for population means based on sample data. The assignment demonstrates the application of statistical methods to solve practical business problems.

STATISTICS FOR BUSINESS
BUS2030
[DATE]

Question 1
Normal distribution
Average spent = $75
Standard deviation =$5
a. Probability that a selected skier spent more than $80 on ski trip
b. Probability that a selected skier spent between $45 and $55
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c. Two values that would fall in middle ninety five percent amount
The z value for 0.975 probability has been found from z table.
The z value = 1.960
Lower value X1 = Mean – (z value * standard deviation) = 75 – (1.960*5) = 65.2
Upper value X2 = Mean + (z value * standard deviation) = 75 + (1.960*5) = 84.8
The two values that would fall in middle 95% amount would be 65.2 and 84.8 respectively.
Question 2
Normal distribution
μ=9 0
σ =1 0
a. X >65
b. X <60
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c. X <70∨X >110
First step: X <70
Second step: X >110
X <70∨X >110=0.0228+0.0228=0.0456
d. The two X values that are 70% of values
The z value for 0.85 probability has been found from z table.
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The z value = 1.036
First value = Mean – (z value * standard deviation) = 90 – (1.036*10) = 79.64
Second value = Mean + (z value * standard deviation) = 90 + (1.036*10) = 100.36
The two X values that are 70% of values are 79.64 and 100.36.
Question 3
Uniform distribution
Lower bound a = 0
Upper bound b = 110
a. Probability that time between arrival of two planes would be less than 40 min
b. Between 60 min and 80 min
c. More than 45 min
Pr ( X >45 ) =110−45
110−0 =0.5909
d. Mean and standard deviation
Mean= ( a+ b )2= ( 0+110 )2=12,100
Standard deviation= b−a
√12 = 110−0
√12 =31.75
Question 4
Mean = 2.63 inches
Standard deviation = 0.03 inches
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Sample size = 9 balls
a. Sampling distribution of mean
Mean = 2.63 inches
Standard erro= Standard deviation
√Sample ¿ ¿= 0.03
√ 9 =0.01¿ ¿
b. Probability that sample mean is less than 2.61 inches
c. Probability that sample mean between 2.62 inches and 2.64 inches
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Question 5
Sample size = 50 bottles
Mean = 0.927 gallon
Standard deviation = 0.04 gallon
95% Confidence interval =?
The z value for 95% confidence interval = 1.96
Lower limit =Mean−Z value∗Standard deviation
√ Sample ¿ ¿=0.927−1.96∗
( 0.04
√ 50 ) =0.916 ¿ ¿
Upper limit =Mean+ Z value∗Standard deviation
√ Sample ¿¿=0.927+ 1.96∗( 0.04
√50 )=0.938 ¿ ¿
95% Confidence interval = [0.92 0.94]
It can be said with 95% confidence that the mean amount of water included in one-gallon
bottle would fall between 0.916 gallon and 0.938 gallon.
Question 6
Sample size = 84
Mean = $82
Standard deviation = $9
95% Confidence interval =?
The t value for 95% confidence interval = 2.0096
Lower limit =Mean−Z value∗Standard deviation
√ Sample ¿ ¿=82−1.96∗
( 9
√ 84 )=80.027 ¿ ¿
Upper limit =Mean+ Z value∗Standard deviation
√ Sample ¿¿=82+1.96∗
( 9
√84 )=83.973 ¿ ¿
95% Confidence interval = [80.027 83.973]
It can be said with 95% confidence that the mean one-time gift donation would fall between
$80.03 and $83.97.
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