Hello! Showing work would be super helpful. I got these questions wrong on a hom

Question

Hello! Showing work would be super helpful. I got these questions wrong on a homework assignment because I am not fully understanding the math behind it. Thanks you very much in advance!

At a certain school, 25% of the students will qualify for financial aid type A and 31% will qualify for financial aid type B. If it is known that 3.5% of the students will qualify for both types of financial aid, find:

1.) The probability that a student wil qualify for financial aid (at least on of either type A or type B)

2.) The probability that a student will qualify for type B given that he or she qualifies for type A.

3.) The probability that a student will not qualify for type B.

4.) Proof that qualifying for the two types of financial aid are not independant events.

Consider the following scenario and calculate the descriptive to answer the accompanying questions:

Clydesdale horses have a mean height of 16.5 hands with a standard deviation of 1.85 hands and Great Danes have a mean height of 81 cm with a standard deviation of 13 cm.

1) Which population exhibits greater variation in height?

2) Which of these two animals is tller compare to its respective breed: a 20.2 hand tall Clydesdale horse or a 107 cm tall Great Dane?

Explanation / Answer

Question 1

Solution:

Let event A be a student will qualify for financial aid type A and event B be a student will qualify for financial aid type B.

We are given

P(A) = 0.25

P(B) = 0.31

P(AB) = 0.035

Part 1

We have to find P(A or B)

P(A or B) = P(AUB)

P(AUB) = P(A) + P(B) – P(AB)

P(AUB) = 0.25 + 0.31 – 0.035

P(AUB) = 0.525

Required probability = 0.525

Part 2

We have to find P(B|A)

P(B|A) = P(AB)/P(A)

P(B|A) = 0.035/0.25

P(B|A) = 0.14

Required probability = 0.14

Part 3

We have to find P(B’)

P(B’) = 1 – P(B)

P(B’) = 1 – 0.31

P(B’) = 0.69

Required probability = 0.69

Part 4

If two events A and B are independent, then P(AB) = P(A)*P(B)

P(AB) = 0.035

P(A)*P(B) = 0.25*0.31 = 0.0775

So, P(AB) P(A)*P(B)

So, events A and B are not independent.

Question 2

Part 1

For Clydesdale horses,

Mean = 16.5

SD = 1.85

Coefficient of variation = C.V. = SD/Mean = 1.85/16.5 = 0.112121 or 11.21%

For Great Danes,

Mean = 81

SD = 13

Coefficient of variation = C.V. = SD/Mean = 13/81 = 0.160494 or 16.05%

Coefficient of variation for Great Danes is greater than coefficient of variation for Clydesdale horses, so population of Great Danes exhibits greater variation in height.

Part 2

We have to find Z scores for given comparison.

Z = (X – mean) / SD

Z for Clydesdale horses

Z = (20.2 – 16.5) / 1.85 = 2.00

Z for Great Danes

Z = (107 – 81) / 13 = 2.00

Z scores for both animals are same.

So, two animals are same in height as compared to its respective bread.

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