Answer:
In both parts of the question, you're asked to find the limit as
n approaches infinity for certain probabilities involving the estimation of the unknown probability of success
π. Given that
=
0.5
π=0.5, we can simplify the expressions and apply limit properties.
Step-by-step explanation:
Let's start with part (a):
a) Find
lim
�
→
∞
�
(
�
(
�
^
−
�
)
≤
�
)
lim
n→∞
P(n(
π
^
−π)≤x), if
�
=
0.5
π=0.5.
In a Bernoulli distribution, the variance of the estimator
�
^
π
^
is given by
Var
(
�
^
)
=
�
(
1
−
�
)
�
Var(
π
^
)=
n
π(1−π)
. Since
�
=
0.5
π=0.5, this variance simplifies to
Var
(
�
^
)
=
1
4
�
Var(
π
^
)=
4n
1
.
We can use the Central Limit Theorem (CLT) here. The CLT states that as
�
n approaches infinity, the distribution of the sample mean approaches a normal distribution with mean
�
μ (population mean) and variance
�
2
�
n
σ
2
, where
�
2
σ
2
is the population variance. Since we have
�
=
0.5
π=0.5 and
Var
(
�
^
)
=
1
4
�
Var(
π
^
)=
4n
1
, we can treat
�
^
π
^
as a sample mean of Bernoulli trials with
�
=
0.5
π=0.5.
Now, let's rewrite the expression
lim
�
→
∞
�
(
�
(
�
^
−
�
)
≤
�
)
lim
n→∞
P(n(
π
^
−π)≤x) as a z-score (standard score) and find the limit:
lim
�
→
∞
�
(
�
(
�
^
−
�
)
Var
(
�
^
)
≤
�
Var
(
�
^
)
)
lim
n→∞
P(
Var(
π
^
)
n(
π
^
−π)
≤
Var(
π
^
)
x
)
Substitute the values:
�
=
0.5
π=0.5 and
Var
(
�
^
)
=
1
4
�
Var(
π
^
)=
4n
1
:
lim
�
→
∞
�
(
2
�
(
�
^
−
0.5
)
1
4
�
≤
�
1
4
�
)
lim
n→∞
P(
4n
1
2n(
π
^
−0.5)
≤
4n
1
x
)
Simplify:
lim
�
→
∞
�
(
4
�
(
�
^
−
0.5
)
≤
2
�
)
lim
n→∞
P(4n(
π
^
−0.5)≤2x)
Notice that the left-hand side now resembles a z-score. As
�
n goes to infinity, the expression will converge to the standard normal distribution's cumulative distribution function (CDF). Therefore, the limit is:
lim
�
→
∞
�
(
4
�
(
�
^
−
0.5
)
≤
2
�
)
=
Φ
(
2
�
)
lim
n→∞
P(4n(
π
^
−0.5)≤2x)=Φ(2x)
where
Φ
Φ represents the standard normal cumulative distribution function.
This limit is not dependent on
�
π and will approach the value of
Φ
(
2
�
)
Φ(2x) as
�
n goes to infinity.
For part (b), the approach is similar, but it involves the logit transformation. The logit transformation of
�
^
π
^
is
logit
(
�
^
)
=
log
(
�
^
1
−
�
^
)
logit(
π
^
)=log(
1−
π
^
π
^
). You would follow a similar process of simplifying and finding the limit as
�
n approaches infinity.
Choose exactly two answers that are correct.
A.
Triangle RST was reflected across the x-axis and then rotated 90° clockwise around the origin.
B.
Triangle RST was reflected across the x-axis and then reflected across the y-axis.
C.
Triangle RST was rotated counterclockwise 90° around the origin and then reflected across the x-axis.
D.
Triangle RST was translated 8 units right and then reflected across the x-axis.
Answer:
16.66%
Step-by-step explanation:
The probability would be the multiplication of the different events.
First event, take out a white rose and plant it, the probability of that is 2/4 = 1/2
Second event, take out a red rose and plant it, the probability is 2/3.
Third event, take out another red rose and plant it, the probability is 1/2.
Now, the probability of everything is:
(1/2) * (2/3) * (1/2) = 0.166
that is to say, the probability of what the statement asks us to happen is 16.66%
Answer:
The answer to the question is
The probability that the 2 rosebushes in the middle of the row will be the red rosebushes = 1/6
Step-by-step explanation:
To solve the question we note that the required permutation is WRRW
Therefore the total number of ways to plant the four rose bushes given that two white rose bushes are in the middle is given by
4!/(2!2!) = 6 Therefore the probability that the 2 rosebushes in the middle of the row will be the red rosebushes = 1/6
Answer:
38.68 degrees
Step-by-step explanation:
In this equation you are trying to find the degrees using the hypotenuse and opposite. So you have to use the function sin^-1(5/8). Because when using Sin it is opposote ove hypotenuse.