Title | Unit 2 Short Answer Section |
---|---|
Author | Adrianne John |
Course | Introduction to Entrepreneurship |
Institution | University of the Cordilleras |
Pages | 3 |
File Size | 162.3 KB |
File Type | |
Total Downloads | 87 |
Total Views | 159 |
Math quiz...
Short Answer U2 1. The shelf life of a particular dairy product is normally distributed with a mean of 12 days and a standard deviation of 3 days. a) Draw the normal curve to represent the normally distributed shelf life of the dairy product. (2 marks)
34%
34%
2.35%
2.35%
0.15%
13.5% 3
6
13.5% 9
12
15
0.15% 18
21
b) What percentage of the dairy product would have a shelf life less than 9 days? (2 marks)
= 13.5 + 2.35 + 0.15 =16% = In less than 9 days 16% of the dairy product would have a shelf life c) What percentage of the dairy product would have a shelf life greater than 18 days? (2 marks)
= 2.35 + 0.15 = 2.50% = 2.5% of the dairy product would have a shelf life greater than 18 days d) What percentage of the dairy product would have a shelf life between 10 and 14 days? (2 marks)
= normalcdf (10, 14, 12, 3) = .495015066 × 100 = 49.5015066 = 49.5% = 49.5% of the dairy product would have a shelf life between 10 and 14 days
2) Research of a manufacturer of LED TV’s shows that the average TV will be 36 months before there is a need of repairs. If the standard deviation for the data is 8 months, what length of time should the manufacturer set for a warranty such that less than 10% of all TV’s will need repairs during the warranty period? (2 marks)
= invnorm ( 0.1 ,36 , 8) = 25.7472 = The manufacturer should offer a 25.75 month warranty =25.75 month warranty
Z=
𝑥− 𝜇 𝜎
-1.2816 =
𝑥−36 8
𝑥−36
(-1.2816)8 = ( 8 )8 -10.2528 = x – 36 -10.2528 + 36 = x 25.7472 = x
3) Convert a normally distributed score of 130 with a mean of 90 and a standard deviation of 14 into a z-score. (2 marks) 𝒙 −𝝁 𝝈
Z= Z=
𝟏𝟑𝟎−𝟗𝟎 𝟏𝟒
=
𝟒𝟎 𝟏𝟒
= 2.857
Z = 2.86
4) For a first year University calculus course the average grade is 62%, with a standard deviation of 10%. If 40 students scored between 69% and 80% on the exam, how many students took the exam? Use z-scores in your solution. (3 marks)
Z= Z=
𝑥−𝑥 𝜎 𝑥−𝑥
Z=
69−62 10
80 −62
Z = 10 𝜎 normalcdf(0.7 ,1.8) 40 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 0.2060
Z = 0.7 Z = 1.8 0.2060
194.17 students took the exam
= 194 students took the exam
5) Determine a 90% confidence interval for µ if σ =5, = 70 and n = 82. Answer interval rounded to two decimal places. (3 marks)
X−
𝒁𝒙 𝟐
𝝈
( )...