Ill give brainliest
2 answers:
You might be interested in
Answer:
99% confidence interval for the population mean is [19.891 , 24.909].
Step-by-step explanation:
We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.
Assuming the population has a normal distribution.
Firstly, the pivotal quantity for 99% confidence interval for the population mean is given by;
P.Q. = ~
where, = sample mean age of selected students = 22.4 years
s = sample standard deviation = 3.8 years
n = sample of students = 19
= population mean
<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>
So, 99% confidence interval for the population mean, is ;
P(-2.878 < < 2.878) = 0.99 {As the critical value of t at 18 degree of
freedom are -2.878 & 2.878 with P = 0.5%}
P(-2.878 < < 2.878) = 0.99
P( <
<
) = 0.99
P( <
<
) = 0.99
<u>99% confidence interval for</u> = [
,
]
= [ ,
]
= [19.891 , 24.909]
Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].
F(x) = (x + 1)(x - 2)
f(x) = x(x - 2) + 1(x - 2)
f(x) = x(x) - x(2) + 1(x) - 1(2)
f(x) = x² - 2x + x - 2
f(x) = x² - x - 2
f(x) = x(x - 2) + 1(x - 2)
f(x) = x(x) - x(2) + 1(x) - 1(2)
f(x) = x² - 2x + x - 2
f(x) = x² - x - 2