Answer:
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 64
Sample mean = 22.3
Sample standard deviation = 8.8
We want to estimate 92% confidence interval.
92% Confidence interval:
Putting the values, we get,
Degree of freedom = n - 1 = 63
Answer:



Step-by-step explanation:

This means:
- when x is equal to zero or less than zero, g(x) will always be 7.
- when x is more than zero, g(x) is




Answer:
-4
Step-by-step explanation:
Given function
f(x) = -2x² + 3x - 4
Comparing with ax² + bx + c We get
c = - 4
Hope it will help :)❤
Answer:
P(2.50 < Xbar < 2.66) = 0.046
Step-by-step explanation:
We are given that Population Mean,
= 2.58 and Standard deviation,
= 0.75
Also, a random sample (n) of 110 households is taken.
Let Xbar = sample mean household size
The z score probability distribution for sample mean is give by;
Z =
~ N(0,1)
So, probability that the sample mean household size is between 2.50 and 2.66 people = P(2.50 < Xbar < 2.66)
P(2.50 < Xbar < 2.66) = P(Xbar < 2.66) - P(Xbar
2.50)
P(Xbar < 2.66) = P(
<
) = P(Z < -1.68) = 1 - P(Z 1.68)
= 1 - 0.95352 = 0.04648
P(Xbar
2.50) = P(
) = P(Z
-3.92) = 1 - P(Z < 3.92)
= 1 - 0.99996 = 0.00004
Therefore, P(2.50 < Xbar < 2.66) = 0.04648 - 0.00004 = 0.046