Given:
n = 50, sample size
, sample mean
s = 2.4 min, sample standard deviation.
The confidence interval is
At the 99% confidence level, t* (from the student's t-distribution) is
t* = 2.68
Therefore
t*(s/√n) = 2.68*(2.4/√50) = 0.9096
The confidence interval is
(23.6-0.9096, 23.6+0.9096) ≈ (22.69, 24.51)
Answer: (22.7, 24.5)
Answer:
(0.806, 0.839)
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of 
.
Suppose we take a poll (random sample) of 3653 students classified as Juniors and find that 3005 of them believe that they will find a job immediately after graduation.
This means that 
99% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
The lower limit of this interval is:
The upper limit of this interval is:
The answer is (0.806, 0.839)
The first step in solving this is to substitute x with 3 and y with 2, so that
becomes
Then simply solve using the order of operations.
First exponents (there are no parentheses)
Then multiply and divide
And finally add and subtract
Therefore the value <span>of 1/3 x^2 + 5.2y when x=3 and y=2 is 13.4</span>
Answer:
0.0159
Step-by-step explanation:
Given that a common practice of airline companies is to sell more tickets for a particular flight than there are seats on the plane, because customers who buy tickets do not always show up for the flight.
Here if X is the no of persons that do not show up, then X is binomial as each trial is independent with p = 0.04 and n =150 (no of tickets sold)
The plane is overbooked if more than 150 show up
i.e. less than 2 do not show up
Hence the probability that the airline overbooked this flight
=
