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zmey [24]
3 years ago
14

The following data were collected by counting the number of operating rooms in use atTampa general Hospital over a 20-day period

: On three of the days only one operatingroom was used, on five of the days two were used, on eight of the days three were used,and on four days all four of the hospital’s operating rooms were used.
a. Use the relative frequency approach to construct an empirical discrete probabilitydistribution for the number of operating rooms in use on any given day.b. Draw a graph of the probability distribution.c. Show that your probability distribution satisfies the required conditions for a validdiscrete probability distribution.

Mathematics
1 answer:
arlik [135]3 years ago
7 0

Answer:

a)

Operating room x    1         2      3       4

P(x)                          0.15  0.25  0.4   0.2

b)

The graph of probability distribution is in attached image.

c)

All the probabilities lies in the range (0 to 1) and the probabilities add up to 1 so, the computed probability distribution is the valid probability distribution.

Step-by-step explanation:

Number of operating rooms   Days

1                                                  3

2                                                 5

3                                                 8

4                                                 4

a)

Sum of frequencies=1+2+3+4=10

Operating rooms x   frequency f  Relative frequency

1                                    3                    3/20=0.15

2                                   5                   5/20=0.25

3                                   8                   8/20=0.4

4                                   4                   4/20=0.2

So, using the relative frequency approach, a discrete probability distribution for number of operating rooms in use on any given day is

Operating room x    1         2      3       4

P(x)                          0.15  0.25  0.4   0.2

b)

The graph of probability distribution is in attached image.

c)

There are two conditions for a probability distribution to be valid

1. All probability must ranges from 0 to 1.

2. All probabilities must add up to 1.

We can see that all the probabilities lies in the range (0 to 1), so, condition 1 is satisfied.

For condition 2,

sum[p(x)]= 0.15+0.25+0.4+0.2=0.4+0.6=1.

As the probabilities add up to 1, so the condition 2 is also satisfied.

Thus, the computed probability distribution is the valid probability distribution.

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Dr. Miriam Johnson has been teaching accounting for over 20 years. From her experience, she knows that 60% of her students do ho
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Answer:

a) The probability that a student will do homework regularly and also pass the course = P(H n P) = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P') = 0.12

c) The two events, pass the course and do homework regularly, aren't mutually exclusive. Check Explanation for reasons why.

d) The two events, pass the course and do homework regularly, aren't independent. Check Explanation for reasons why.

Step-by-step explanation:

Let the event that a student does homework regularly be H.

The event that a student passes the course be P.

- 60% of her students do homework regularly

P(H) = 60% = 0.60

- 95% of the students who do their homework regularly generally pass the course

P(P|H) = 95% = 0.95

- She also knows that 85% of her students pass the course.

P(P) = 85% = 0.85

a) The probability that a student will do homework regularly and also pass the course = P(H n P)

The conditional probability of A occurring given that B has occurred, P(A|B), is given as

P(A|B) = P(A n B) ÷ P(B)

And we can write that

P(A n B) = P(A|B) × P(B)

Hence,

P(H n P) = P(P n H) = P(P|H) × P(H) = 0.95 × 0.60 = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P')

From Sets Theory,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

P(H n P) = 0.57 (from (a))

Note also that

P(H) = P(H n P') + P(H n P) (since the events P and P' are mutually exclusive)

0.60 = P(H n P') + 0.57

P(H n P') = 0.60 - 0.57

Also

P(P) = P(H' n P) + P(H n P) (since the events H and H' are mutually exclusive)

0.85 = P(H' n P) + 0.57

P(H' n P) = 0.85 - 0.57 = 0.28

So,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

Becomes

0.03 + 0.28 + 0.57 + P(H' n P') = 1

P(H' n P') = 1 - 0.03 - 0.57 - 0.28 = 0.12

c) Are the events "pass the course" and "do homework regularly" mutually exclusive? Explain.

Two events are said to be mutually exclusive if the two events cannot take place at the same time. The mathematical statement used to confirm the mutual exclusivity of two events A and B is that if A and B are mutually exclusive,

P(A n B) = 0.

But, P(H n P) has been calculated to be 0.57, P(H n P) = 0.57 ≠ 0.

Hence, the two events aren't mutually exclusive.

d. Are the events "pass the course" and "do homework regularly" independent? Explain

Two events are said to be independent of the probabilty of one occurring dowant depend on the probability of the other one occurring. It sis proven mathematically that two events A and B are independent when

P(A|B) = P(A)

P(B|A) = P(B)

P(A n B) = P(A) × P(B)

To check if the events pass the course and do homework regularly are mutually exclusive now.

P(P|H) = 0.95

P(P) = 0.85

P(H|P) = P(P n H) ÷ P(P) = 0.57 ÷ 0.85 = 0.671

P(H) = 0.60

P(H n P) = P(P n H)

P(P|H) = 0.95 ≠ 0.85 = P(P)

P(H|P) = 0.671 ≠ 0.60 = P(H)

P(P)×P(H) = 0.85 × 0.60 = 0.51 ≠ 0.57 = P(P n H)

None of the conditions is satisfied, hence, we can conclude that the two events are not independent.

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