Statisticians for a roadside assistance company interviewed 50,000 randomly selected United States (US) households. Of those, 15

Response:

a) S ~ N (0, 48)

b) P(S > 10) = 0.0745

Detailed explanation:

Given Information:-

- Total number of questions, n = 100

- Each question has 5 options

- The probability of correctly guessing each answer is independent.

- Points for a correct answer = +4

- Points for an incorrect answer = -1

Inquiries:-

a) Determine????(S).

b) Determine P(S>10). Represent your response as a mathematical formula, then utilize the code cell below to calculate its numerical value, providing both the calculation and its result.

Solution:-

- The probability (p) for answering a question correctly is:

p (correct answer) = 1/5 = 0.2

- The expected number of correct and incorrect answers can be calculated as follows:

(Expected correct answers) = n*p = 100*0.2 = 20

(Expected incorrect answers) = n*(1-p) = 100*0.8 = 80

- The anticipated score for correct answers will be:

Sc(u) = (Points for a correct answer)*(Expected correct answers)

Sc(u) = (+4)*(20)

Sc(u) = 80 points

The anticipated score for incorrect answers will be:

Si(u) = (Points for an incorrect answer)*(Expected incorrect answers)

Si(u) = (-1)*(80)

Si(u) = -80 points.

- The average score a student might achieve would be S(u):

S(u) = Sc(u) + Si(u)

S(u) = 80 - 80 = 0

- The variance for both correct and incorrect answers can be calculated as:

Var(correct answers) = n*p*q = 100*0.2*0.8 = 16

Var(incorrect answers) = n*p*q = 100*0.2*0.8 = 16

- The variance of points for correct answers can be expressed as:

Sc(Var) = Var(correct answer) * (Points for a correct answer)

Sc(Var) = 16*(+4) = +64 points

- The variance of points for incorrect answers can be expressed as:

Si(Var) = Var(incorrect answer) * (Points for an incorrect answer)

Si(Var) = 16*(-1) = -16 points

- Since the probabilities of correct guesses are independent, according to the independence principle:

S(Var) = Sc(Var) + Si(Var)

= 64 - 16

= +48 points

- The standard deviation for the score distribution (s.d) is:

S(s.d) = √S(Var) = √48 = 6.9282

- Therefore, the anticipated score (S) from guessing on the MCQ test would yield a mean of u = 0 points and s.d = + 48 points.

- The random variable (S) can be approximated using normal distribution as follows:

S ~ N (0, 48)

- To find the required probability P(S>10).

Calculate the Z-value for S = 10 points:

Z-value =  ( S - u ) / s.d

=  ( 10 - 0 ) / 6.9282

= 1.4434

Consult the standardized Z-table for normal distribution:

P(Z > 1.4434) = 0.0745

The probability is:

P(S > 10) = P(Z > 1.4434) = 0.0745

Answer:

A. P = 0.73

B. P(A∩B∩C') = 0.22

C. P(B/A) = 0.5

P(A/B) = 0.75

D. P(A∩B/C) = 0.4

E. P(A∪B/C) = 0.85

Step-by-step explanation:

Denote A as the event of a student having a Visa card, B as the event of holding a MasterCard, and C as the event of owning an American Express card. Additionally, let A' indicate the event of not having a Visa card, B' signify not having a MasterCard, and C denote the event of not possessing an American Express card.

Thus, with the given probabilities, we can determine the following probabilities:

P(A∩B∩C') = P(A∩B) - P(A∩B∩C) = 0.3 - 0.08 = 0.22

Here, P(A∩B∩C') refers to the chance that a student has both a Visa and MasterCard but does not own an American Express, P(A∩B) indicates the probability that a student possesses both a Visa and a MasterCard, and P(A∩B∩C) represents the likelihood that a student has a Visa, MasterCard, and American Express. Similarly, we can find:

P(A∩C∩B') = P(A∩C) - P(A∩B∩C) = 0.15 - 0.08 = 0.07

P(B∩C∩A') = P(B∩C) - P(A∩B∩C) = 0.1 - 0.08 = 0.02

P(A∩B'∩C') = P(A) - P(A∩B∩C') - P(A∩C∩B') - P(A∩B∩C)

                   = 0.6 - 0.22 - 0.07 - 0.08 = 0.23

P(B∩A'∩C') = P(B) - P(A∩B∩C') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.4 - 0.22 - 0.02 - 0.08 = 0.08

P(C∩A'∩A') = P(C) - P(A∩C∩B') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.2 - 0.07 - 0.02 - 0.08 = 0.03

A. The likelihood that the selected student holds at least one of the three card types is calculated as follows:

P = P(A∩B∩C) + P(A∩B∩C') + P(A∩C∩B') + P(B∩C∩A') + P(A∩B'∩C') +              

     P(B∩A'∩C') + P(C∩A'∩A')

P = 0.08 + 0.22 + 0.07 + 0.02 + 0.23 + 0.08 + 0.03 = 0.73

B. The probability that the chosen student possesses both a Visa and a MasterCard without an American Express card can be represented as P(A∩B∩C') equaling 0.22

C. P(B/A) represents the chance that a student holds a MasterCard provided they have a Visa. This is calculated as:

P(B/A) = P(A∩B)/P(A)

By substituting in the values, we find:

P(B/A) = 0.3/0.6 = 0.5

In a similar manner, P(A/B) represents the probability a student has a Visa given they possess a MasterCard, calculated as:

P(A/B) = P(A∩B)/P(B) = 0.3/0.4 = 0.75

D. For a student with an American Express card, the likelihood they also hold both a Visa and a MasterCard is expressed as P(A∩B/C), calculated as:

P(A∩B/C) = P(A∩B∩C)/P(C) = 0.08/0.2 = 0.4

E. If the student has an American Express card, the probability they possess at least one of the other two card types is denoted as P(A∪B/C), computed as:

P(A∪B/C) = P(A∪B∩C)/P(C)

Where P(A∪B∩C) = P(A∩B∩C)+P(B∩C∩A')+P(A∩C∩B')

Consequently, P(A∪B∩C) equals 0.08 + 0.07 + 0.02 = 0.17

Ultimately, P(A∪B/C) equals:

P(A∪B/C) = 0.17/0.2 =0.85