Investing All Your Money in a Few Randomly Chosen Stocks That Make Up the Index

Ex. 4

a) Investing all your money in a few randomly chosen stocks that make up the index

b) Investing all your money in a few randomly chosen stocks that make up the index

Ex. 22

1. Assuming proportion as .5 or 50%. P=.5

Standard error = 1.58%

Margin of error =2 x

Standard error =2*.0158= 3.16%

1. For margin of error as 10% or .1

Standard error =10%/2 =5%

.05 =sqrt(.5*.5/n) N=100

The poll sample of 100 is required.

1. For a proportion as 48% or .48

As per my calculations shown below. I would around get 1% or 0.01. This is the Standard Error for a percentage. With a genuine percentage of 48%, The Standard Error for Percentage gives 0.01 or 1%.

48% x(100%-48%) = 0.2496.

0.2496/2500 =0.00009984

SE = SQ(0.00009984) = 0.01

ME = 0.01 X 2 = 2%

Ex. 28

Margin of error =.5% =.005

Standard deviation =2.5% =.025

Assuming the confidence interval as 95% =1.96.

(1.96 x .025/ sqrt)/0.05

(0.98)2 = 0.9604

= 1

Therefore, required sample size is 1.

Ex. 17

1. Proportion =145/1100 =.1318

Standard error=sqrt(.1318 x (1-.1318)/1100)=.0102

Margin of error =.0102 x 1.96 = .02

Therefore,

 Lower limit =.1318-.02 =.1118=11.18%

Upper limit =.1318+.02 =.1518=15.18%

b) 95% sure that between 11.18% to 15.18% workers at a company accessed Internet sites that were clearly not identified with their occupations in the most recent week.

c) 10% is excluded in the confidence interval (11.18%-15.18%) consequently the organization statistics is not the same as the national insights.

 d) Yes, 95% of individuals at this organization spent in the scope of 4.8 hours give or take 4.2 hours on non-work related destinations on the grounds that the 95% certainty interim is around 4.8+\-1.96*2.1 (4.8+\-4.2)

Ex. 20

a) For a flight with 400 seats, the airline can be 95% confident that between 87.06% and

92.94% percent of the people will actually show up.

p=.9

[pic 1] where p = 0.90, [pic 2]= 1.959963985, n = 400

[pic 3]

= (0.8706, 0.9294)

b) The approximate number of tickets should they sell so that there is less than a 10% chance they run out of seats can be calculated by trial and error method.

We know that there 400 seats in the flight and only about 90% of the people who but the ticket actually show up.

Let X be the number of tickets actually show up. Clearly X is binomial with n = 400 and p = 0.90.  The probability mass function binomial variable is given by[pic 4].