Answer:
95% confidence interval for the percentage of all U.S. public high school students who are obese is [0.110 , 0.190].
Step-by-step explanation:
We are given that 15% of a random sample of 300 U.S. public high school students were obese.
Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;
P.Q. =
~ N(0,1)
where,
= sample % of U.S. public high school students who were obese = 15%
n = sample of U.S. public high school students = 300
p = population percentage of all U.S. public high school students
<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>
<u></u>
<u>So, 95% confidence interval for the population proportion, p is ;</u>
P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level
of significance are -1.96 & 1.96}
P(-1.96 <
< 1.96) = 0.95
P(
<
<
) = 0.95
P(
< p <
) = 0.95
<u>95% confidence interval for p</u> = [
,
]
= [
,
]
= [0.110 , 0.190]
Therefore, 95% confidence interval for the percentage of all U.S. public high school students who are obese is [0.110 , 0.190].
Answer:
I like the the answer so I think that it is C
Answer:
a) 0.1108
(b) 0.0173
Step-by-step explanation:
We are given that 20% of all stock investors are retired people. A random sample of 25 stock investors is taken.
Firstly, the binomial probability is given by;
where, n = number of trails(samples) taken = 25
r = number of successes
p = probability of success and success in our question is % of
retired people i.e. 20%.
Let X = Number of people retired
(a) Probability that exactly seven are retired people = P(X = 7)
P(X = 7) =
= = 0.1108
(b) Probability that 10 or more are retired people = P(X >= 10)
P(X >= 10) = 1 - P(X <= 9)
Now, using binomial probability table, we find that P(X <= 9) is 0.98266 at n = 25, p = 0.2 and x= 9
So, P(X >= 10) = 1 - 0.98266 = 0.0173.
Answer:
15
Step-by-step explanation:
you put -5 in place of the x the you multiply 5 times -5
then you subtract that from 40
so 40 - 25