From Socr
Answer:
a. false, the standard deviation of the sample mean is
![\frac{\sigma}{\sqrt n}](/socr/uploads/math/8/0/8/8086b0bb03c0f90e93a7fb357971bc7e.png)
. Thus as the sample size increases, n increases, and as n increases, the standard deviation decreases.
b. True
c. False, standard deviation of the sample mean is
d. True, the standard deviation of the total of a sample of n observations is
![n\sqrt \sigma](/socr/uploads/math/6/e/5/6e58325f825a239b6ff49601f38ac126.png)
; but the standard deviation of the sample mean is
![\frac{\sigma}{\sqrt n}.](/socr/uploads/math/a/a/9/aa955b9520ceeaaab0a4a5523e018656.png)
Unless n is one, the standard deviation of the total of a sample of n observations exceeds the standard deviation of the sample mean.
e. False, let's assume
σ = 2 and
n = 2. In this case, the z-score for
![P(\overline{X} > 4)](/socr/uploads/math/8/0/0/800221cf9269f8029e1fc97db890f1ec.png)
would be -2.828 while the z-score for
P(X > 4) would be -2.
P(Z > − 2.828) > P(Z > − 2). Therefore the statement is false.
Answer:
a.
Failed to parse (syntax error): P(X \ge 1000)= P(X=1000)+P(X=1001)+....+P(X=1500)= (1500 \choose 1000) \times (.7)^1000 \times (.3)^500 + (1500 \choose 1001) \times (.7)^1001 \times (.3)^499 + ...+(1500 \choose 1500) \times (.7)^1500 \times (.3)^0 = \summation (\1500 \choose X) \times (.7)^X \times (.3)^1500-X
b. We can use the normal approximation to binomial:
![\mu = np = 1500 \times 0.70 = 1050.](/socr/uploads/math/5/c/6/5c67258b137153bdd7d166630b863d47.png)
and
Below you can see a snapshot for this approximation:
Answer:
a.
Below you can see a snapshot for this part:
b. ??
c. The central limit theorem states that the sample mean approaches the normal distribution as the sample size gets bigger. Usually, if
![n \ge 30](/socr/uploads/math/c/7/8/c78e8c182021e831feadef82b09e861f.png)
we can assume that the sample mean approaches the normal distribution. In this case
n = 400. Therefore n satisfies the requirement of a large n.
d.
![\overline{X} \sim N(8,1).](/socr/uploads/math/1/6/0/160d583e2a7372ca34eb3ebe12c17ec2.png)
According to the snapshot below, the middle 80% of this distribution is (6.721,9.279). Therefore
w = 8 − 6.721 = 1.29
e.
![T \sim N(n\mu,\sigma\sqrt n).](/socr/uploads/math/3/b/5/3b567a35a6f91b1315298f1ca341019b.png)
In this case,
![T \sim N(3200,400).](/socr/uploads/math/8/a/c/8ac3b48c50a8992963f71ac07081fc6e.png)
We know that
P(T > b) = .975.So now we need to find the 97.5th percentile of this distribution using SOCR. According to the SOCR snapshot below, the 97.5th percentile of this distribution is 3984. Therefore b=3984.
Answer:
a.
![\overline{X} \sim N(\mu, \frac{\sigma}{\sqrt n }](/socr/uploads/math/3/3/5/3352415ac086e4d92daba8ab7955cabd.png)
. In this case,
![\overline{X} \sim N(80000, 4518.48)](/socr/uploads/math/3/0/7/307b3ce4446b8aeefad9e9502d500d12.png)
.
b. We can find the answer using SOCR. The answer is 0.004032. Please see snapshot below:
c. We can find the answer right away using SOCR. Please see snapshots below:
This is the distribution for X
This is the distribution for
The probabilities are 55.6% for one hour vs. 86.6% for sample mean. Therefore the sample mean is more likely to be greater than 75000 hours.
Answer:
a. According to the SOCR snapshot below, the 75th percentile is 0.006115.
b.
P(X > .01) = .13.We can see this in the snapshot below:
c.
i.
ii.
iii.One day's return is more likely to be greater than .007. The probabilities are 0.21 for
X vs. .00022 for
![\overline{X}](/socr/uploads/math/f/7/9/f7991af03fae085812323528a0605b4b.png)
.
This is the snapshot for P(X > .007)
This is the snapshot for