Answer:
And we can find this probability using the complement rule and the normal standard table or excel:
The firgure attached illustrate the problem
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the retirement savings of a population, and for this case we know the distribution for X is given by:
Where
and
We are interested on this probability
And the best way to solve this problem is using the normal standard distribution and the z score given by:
If we apply this formula to our probability we got this:
And we can find this probability using the complement rule and the normal standard table or excel:
The firgure attached illustrate the problem
Answer:
de
Step-by-step explanation:
Step-by-step explanation:
-5 + 2 - 8 - (-14) =
(-3) - 8 + 14 =
-11 + 14 =
3 =
not sure but here you go^^
2(3t - 1) - 5(t - 1)
2(3t) - 2(1) - 5(t) + 5(1)
6t - 2 - 5t + 5
6t - 5t - 2 + 5
t + 3
Step-by-step explanation:
Mean = 81740
Standard deviation = 4590
Sample size = 15
Alpha level = 1-0.95 = 0.05
Df = 15-1 = 14
Critical value:
alpha/2 = 0.05/2 = 0.05
t0.025
t critical value = 2.145
Margin of error ME
2.145 x 4590/√15
2.145 x 4590/3.873
ME = 2542.09
Confidence interval
Lower CI = mean - ME
= 81740-2542.09
= 79197.91
Upper CI = mean + ME
= 81740+2542.09
= 84282.09
[ 79197.91, 84282.09]
B.
Using excel, exact answer for CI
Lower limit = 79198.142724212173
Upper limit = 84281.8572757827
C.
The assumptions to be made from the population are that
1. Samples are random
2. These samples are gotten from an approximately normal distribution