Describe 59 on two other ways

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

The average speed for his entire journey from York to Blackpool is about 61.41 km/h.

Here’s a breakdown of how we arrive at this:

The distance he travelled from York to Leeds is 45 km,

and the speed during that section was 54 km/h.

Therefore, the time taken to travel from York to Leeds is 45/54 hours (since Time = Distance/Speed).

Next, the distance from Leeds to Blackpool is 42 km,

and the time for that leg of the journey is 35 minutes, which is 35/60 hours.

This leads to the total duration for his trip as

hours.

The cumulative distance covered equals 45 + 42 = 87 km.

Thus, his average speed is calculated as:

Answer:

a. Alpha equals 3.014 while beta equals 12.442

b. The likelihood that the data transfer duration surpasses 50ms is 0.238

c. The chance that data transfer time falls between 50 and 75 ms is 0.176

Step-by-step explanation:

a. Given the data, the mean and standard deviation for the random variable X are 37.5 ms and 21.6, respectively.

Thus, E(X)=37.5 and V(X)=(21.6)∧2  

To find alpha, we need to apply the formula:

alpha=E(X)∧2/V(X)

alpha=(37.5)∧2/21.6∧2

alpha=1,406.25 /466.56

​alpha=3.014

To determine beta, the following formula is employed:

β=  V(X) ∧2/E(X)

β=(21.6)  ∧2/37.5

β=466.56 /37.5

β=12.442

b. With E(X)=37.5 and V(X)=(21.6)∧2,  

Hence, P(X>50)=1−P(X≤50)

To find the probability of data transfer time exceeding 50ms, we use the formula:

P(X>50)=1−P(X≤50)

=1−0.762

=0.238

The chance of data transfer time exceeding 50ms is 0.238

c. With E(X)=37.5 and V(X)=(21.6)∧2,  

Thus, P(50<X<75)=P(X<75)−P(X<50)  

To find the probability that data transfer time is between 50 and 75 ms, we apply the formula:

P(50<X<75)=P(X<75)−P(X<50)

=0.938−0.762

=0.176

​

The probability that data transfer time falls between 50 and 75 ms is 0.176