The first thing we need to do is check to see if the distribution is approximately normal. np = 50(0.6) = 30 and ng = 50(0.4) 50(0.4) = 20 = Since np 10 and nq ≥ 10, we can conclude that p is approximately a normal distribution with • mean μ = 0.6 and p ● standard deviation бр = pq n = = 0.6(0.4) 50 pq n op nearest thousandth). = V (which is very close to what we saw in our simulation, standard error =0.063). = 0.24 50 Now find the approximation for samples of size 250. Is the standard deviation close to the standard error found in the simulation (standard error = 0.030)? - 0.0693 (Round to the Is the standard deviation close to the standard deviation (standard error) found
The first thing we need to do is check to see if the distribution is approximately normal. np = 50(0.6) = 30 and ng = 50(0.4) 50(0.4) = 20 = Since np 10 and nq ≥ 10, we can conclude that p is approximately a normal distribution with • mean μ = 0.6 and p ● standard deviation бр = pq n = = 0.6(0.4) 50 pq n op nearest thousandth). = V (which is very close to what we saw in our simulation, standard error =0.063). = 0.24 50 Now find the approximation for samples of size 250. Is the standard deviation close to the standard error found in the simulation (standard error = 0.030)? - 0.0693 (Round to the Is the standard deviation close to the standard deviation (standard error) found
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
![The first thing we need to do is check to see if
the distribution is approximately normal.
50(0.6) = 30 and
ng = 50(0.4) = 20
np =
пр
Since np 10 and nq ≥ 10, we can
conclude that is approximately a normal
distribution with
• mean Up
• standard deviation
op
=
=
0.6 and
pq
n
=
=
0.6(0.4)
50
=
pq
n
0.24
50
(which is very close to what we saw in our
simulation, standard error =0.063).
standard error found in the simulation
(standard error = 0.030)?
ор
nearest thousandth).
Now find the approximation for samples of size
250. Is the standard deviation close to the
=
0.0693
(Round to the
Is the standard deviation close to the
standard deviation (standard error) found
in the simulation?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F96596ae2-eb5d-4a8c-b967-0ad103142e62%2Fdff73b8c-206a-474f-98af-8755d7012a60%2F452v7va_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The first thing we need to do is check to see if
the distribution is approximately normal.
50(0.6) = 30 and
ng = 50(0.4) = 20
np =
пр
Since np 10 and nq ≥ 10, we can
conclude that is approximately a normal
distribution with
• mean Up
• standard deviation
op
=
=
0.6 and
pq
n
=
=
0.6(0.4)
50
=
pq
n
0.24
50
(which is very close to what we saw in our
simulation, standard error =0.063).
standard error found in the simulation
(standard error = 0.030)?
ор
nearest thousandth).
Now find the approximation for samples of size
250. Is the standard deviation close to the
=
0.0693
(Round to the
Is the standard deviation close to the
standard deviation (standard error) found
in the simulation?
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