A survey finds that 40% of students ride in a car to school. Eri would like to estimate the probability that if 3 students were
2 answers:
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Answer:
Z-score 1.96
Margin of error d= 0.04323
Step-by-step explanation:
Hello!
The study variable in this case is
X: Number of people that oppose the use of photo-cop for issuing traffic tickets in a sample of 500.
n= 500
The parameter of interest is the proportion of people that opposes it.
The estimated proportion is p'= 300/500= 0.6
To estimate the population proportion per Confidence interval you have to approximate the distribution of the sample proportion p' to normal:
p'≈N(p;p(1-p)1/n)
The mean of this distribution is p and the variance is p*(1-p)*1/n
To construct the Confidence interval, since the value of the population proportion is unknown, an estimated variance is used:
p'(1-p')*1/n ⇒ then the estimated standard error is √(p'(1-p')*1/n)
The formula for the confidence interval is:
p' ± *
Where " *
" represents the margin of error of the interval.
Now for a confidence level of 0.95 the value of Z is
The estimated standard error is already calculated:
The margin of error (d) is then:
d= *
= 1.965 * 0.022
d= 0.04323
I hope it helps!
Answer:
The points (x,y,z) that respond to Ir-r0I =1, are all that describes the form with:
-1+x₀<x<1+x₀
-1+y₀<y<1+y₀
-1+z₀<z<1+z₀
Step-by-step explanation:
All points required in this problem came from applying the definition of modulus of a vector:
Ir-r0I =1.