SOLUTION: #1 (5 pts) 1. Forty percent of the Bowie State faculty teach online classes and 25% of the faculty teach part time. Of the part time faculty, 60% have taught an online class. Let

Algebra ->  Probability-and-statistics -> SOLUTION: #1 (5 pts) 1. Forty percent of the Bowie State faculty teach online classes and 25% of the faculty teach part time. Of the part time faculty, 60% have taught an online class. Let       Log On


   



Question 1048856: #1
(5 pts) 1. Forty percent of the Bowie State faculty teach online classes and 25% of the faculty teach part time. Of the part time faculty, 60% have taught an online class. Let A = event that a Bowie State faculty member teaches an online class and B = event that a UMUC faculty member is part time.
a. Find P(A AND B).
b. Find P(B|A).
c. Find P(A OR B).
d. Using an appropriate test, show whether A and B are independent.
e. Using an appropriate test, show whether A and B are mutually exclusive.
Answer:

Work:
#2
(5 pts) Height and weight are two measurements used to track a child’s development. The World Health Organization measures child development by comparing the weights of children who are the same height and the same gender. In 2015, weights for all 95 cm girls in the reference population had a mean μ = 12.7 kg and standard deviation σ = 1.1 kg. Weights are normally distributed. Calculate the z-scores that correspond to the following weights and interpret them.
a. 13.1 kg
b. 8.5 kg
c. 11.2 kg

Answer:
Work:
(13.1-12.7)/1.1=0.4/0.8=0.363
13.1kg is 0.363 SD’s above the value average
(8.5-12.7)/1.1=(-4.2)/1.1=-3.818
8.5kg is below value average by 3.818 SD’s
(11.2-12.7)/1.1=(-1.5)/1.1=-1.363
8.5kg is below value average by 1.363 SD’s


#3
(5 pts) In 2015, the average length of Hillary Clinton’s emails were 174 words, with a standard deviation of 55 words. Suppose we randomly pick 28 of her emails from 2015.
a. In words, Χ = _____________
b. In words, X ¯ = _____________
c. X ¯ ~ _____(_____,_____)
d. The IQR for X ¯ =
Answer:
In words, X= average length of Clinton’s emails
In Words, X ¯= length of 28 randomly picked emails
X ¯ ~ _____(_____,_____)
Work:
#4
(5 pts) The average length of a maternity stay in a U.S. hospital is said to be 2.4 days with a standard deviation of 0.9 days. We randomly survey 80 women who recently bore children in a U.S. hospital.
a. In words, X = _____________
b. In words, X ¯ = ___________________
c. X ¯ ~ _____(_____,_____)
d. In words, ΣX = _______________
e. ΣX ~ _____(_____,_____)
f. Is it likely that an individual stayed more than five days in the hospital? Why or why not?
g. Is it likely that the average stay for the 80 women was more than five days? Why or why not?
h. Which is more likely:
1. An individual stayed more than five days.
2. The average stay of 80 women was more than five days.
i. If we were to sum up the women’s stays, is it likely that, collectively they spent more than a year in the hospital?
Answer:
X= average length of maternity stay in US
X ¯ =
N(2.4,0.9/√80)
In words ΣX =
Work:

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
Suggestions:
2) z =
a) .364, b) -3.818, c) -1.364
I. Round properly according to decimal points listed
II. May want to use z-scores for giving % of the Area to the left of that
particular z-score under the standard normal curve for interpretation purposes
For ex: z = .364, 64.2% Percentile rating (Use Calculator or table)
In Question 3 & 4
x represents the values found from the random samples
(length found for of each of those e-mails sampled, for ex, as in Q-3
x̄ = sample mean = Σx/n

Hope this helps