A professor's son, having made the wise decision to drop out of college, has been finding his way in life taking one job or anot
2 answers:
Answer:
0.9641 or 96.41%
Step-by-step explanation:
Mean career duration (μ) = 88 weeks
Standard deviation (σ) = 20
The z-score for any given career duration 'X' is defined as:
In this problem, we want to know what is the probability that the professor's son's next career lasts more than a year. Assuming that a year has 52 weeks, the equivalent z-score for a 1-year career is:
According to a z-score table, a z-score of -1.8 is at the 3.59-th percentile, therefore, the likelihood that this career lasts more than a year is given by:
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Answer:
We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.
Step-by-step explanation:
The null Hypothesis: Geographical distribution of hotline callers could be the same as the U.S. population distribution
Alternative hypothesis: Geographical distribution of hotline callers could not be the same as the U.S. population distribution
The populations considered are the Midwest, South, Northeast, and west.
The number of categories, k = 4
Number of recent calls = 200
Let the number of estimated parameters that must be estimated, m = 0
The degree of freedom is given by the formula:
df = k - 1-m
df = 4 -1 - 0 = 3
Let the significance level be, α = 5% = 0.05
For α = 0.05, and df = 3,
from the chi square distribution table, the critical value = 7.815
<u>Observed and expected frequencies of calls for each of the region:</u>
<u>Northeast</u>
Observed frequency = 39
It contains 18.1% of the US Population
The probability = 0.181
Expected frequency of call = 0.181 * 200 = 36.2
<u>Midwest</u>
Observed frequency = 55
It contains 21.9% of the US Population
The probability = 0.219
Expected frequency of call = 0.219 * 200 =43.8
<u>South</u>
Observed frequency = 60
It contains 36.7% of the US Population
The probability = 0.367
Expected frequency of call = 0.367 * 200 = 73.4
<u>West</u>
Observed frequency = 46
It contains 23.3% of the US Population
The probability = 0.233
Expected frequency of call = 0.233 * 200 = 46
Where observed frequency
Expected frequency
Calculate the test statistic value, x²
Since the test statistic value, x²= 5.535 is less than the critical value = 7.815, the null hypothesis will not be rejected, i.e. it will be accepted. We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.
Answer:
(a) 0.00605
(b) 0.0403
(c) 0.9536
(d) 0.98809
Step-by-step explanation:
We are given that 40% of first-round appeals were successful (The Wall Street Journal, October 22, 2012) and suppose ten first-round appeals have just been received by a Medicare appeals office.
This situation can be represented through Binomial distribution as;
where, n = number of trials (samples) taken = 10
r = number of success
p = probability of success which in our question is % of first-round
appeals that were successful, i.e.; 40%
So, here X ~
(a) Probability that none of the appeals will be successful = P(X = 0)
P(X = 0) =
= = 0.00605
(b) Probability that exactly one of the appeals will be successful = P(X = 1)
P(X = 1) =
= = 0.0403
(c) Probability that at least two of the appeals will be successful = P(X>=2)
P(X >= 2) = 1 - P(X = 0) - P(X = 1)
= 1 -
= 1 - 0.00605 - 0.0403 = 0.9536
(d) Probability that more than half of the appeals will be successful = P(X > 0.5)
For this probability we will convert our distribution into normal such that;
X ~ N()
and standard normal z has distribution as;
Z = ~ N(0,1)
P(X > 0.5) = P( >
) = P(Z > -2.26) = P(Z < 2.26) = 0.98809