What are 5 numbers that have 5, 11, and 2 as prime factors
1 answer:
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Answer:
[0.7210,0.8189] = [72.10%, 81.89%]
Step-by-step explanation:
The sample size is
n = 200
the proportion is
p = 154/200 = 0.77
<em>Since both np ≥ 10 and n(1-p) ≥ 10 </em>
<em>We can approximate this discrete binomial distribution with the continuous Normal distribution. As the sample size is large enough, not applying the continuity correction factor makes no significant difference. </em>
The approximation would be to a Normal curve with this parameters:
<em>Mean </em>
p = 0.77
<em>Standard deviation </em>
The 90% confidence interval for the proportion would be then
where is the 10% critical value for the Normal N(0,1) distribution , this is a value such that the area under the N(0,1) curve outside the interval
is 10%=0.1
We can either use a table, a calculator or a spreadsheet to get this value.
In Excel or OpenOffice Calc we use the function
<em>NORMSINV(0.95) and we get a value of 1.645 </em>
The 90% confidence interval for the proportion is then
This means there is a 90% probability that the proportion of people who watch educational television is between 72.10% and 81.89%
If the television company wanted to publicize the proportion of viewers, do you think it should use the 90% confidence interval?
Yes, I do.
Substituting <em>x</em> = 0 directly gives the indeterminate form 0/0, so you can use l'Hopital's rule (and you'll have to do that twice):
The limand is continuous everywhere, so you can plug in <em>x</em> = 0 to get 36•1/2 = 18.