Respuesta :
Answer:
a) P(x ≤ 60) = 0.0087
b) P(60 < x < 75) = 0.3009
c) About 6 students will be unable to finish the exam in the allotted time
Step-by-step explanation:
This is a normal distribution problem with
Mean = μ = 79 minutes
Standard deviation = σ = 8 minutes.
a) The probability of completing the exam in one hour or less. P(x ≤ 60)
We first standardize/normalize 60 minutes
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (60 - 79)/8 = - 2.38
To determine the probability of completing the exam in one hour or less.
P(x ≤ 60) = P(z ≤ -2.38)
We'll use data from the normal probability table for these probabilities
P(x ≤ 60) = P(z ≤ -2.38) = 0.00866 = 0.0087 to 4d.p
b) The probability that a student will complete the exam in more than 60 minutes but less than 75 minutes. P(60 < x < 75)
We first standardize/normalize 60 and 75 minutes
For 60 minutes
z = (x - μ)/σ = (60 - 79)/8 = - 2.38
For 75 minutes
z = (x - μ)/σ = (75 - 79)/8 = - 0.50
To determine the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes.
P(60 < x < 75) = P(-2.38 < z < -0.50)
We'll use data from the normal probability table for these probabilities
P(60 < x < 75) = P(-2.38 < z < -0.50)
= P(z < -0.50) - P(z < -2.38)
= 0.30954 - 0.00866
= 0.30088 = 0.3009
c. Assume that the class has 60 students and that the examination period is 90 minutes in length. How many students do you expect will be unable to complete the exam in the allotted time?
Probability of finishing in 90 minutes or less = P(x ≤ 90)
We first standardize/normalize 90 minutes
z = (x - μ)/σ = (90 - 79)/8 = 1.38
To determine the probability of completing the exam in 90 minutes or less.
P(x ≤ 90) = P(z ≤ 1.38)
We'll use data from the normal probability table for these probabilities
P(x ≤ 90) = P(z ≤ 1.38) = 0.91621
This means 91.621% of the class will finish just in time.
91.621% × 60 = 54.97 ≈ 54 (we can't round the number up because it represents number of students that finish in time and rounding a 0.97 to a 1 makes the analysis wrong)
About 54 students will finish in time.
Meaning about 6 students will be unable to finish the exam in the allowed time.
Hope this Helps!!!
a) The probablity will be,P(x ≤ 60) = 0.0087
b) P(60 < x < 75) = 0.3009
c) About 6 students will be unable to finish the exam in the allotted time
Calculation of probability:
This is a normal distribution problem with
Mean = μ = 79 minutes
Standard deviation = σ = 8 minutes.
a) The probability of completing the exam in one hour or less. P(x ≤ 60)
We first standardize in 60 minutes
z = (x - μ)/σ = (60 - 79)/8 = - 2.38
To determine the probability of completing the exam in one hour or less.
P(x ≤ 60) = P(z ≤ -2.38)
P(x ≤ 60) = P(z ≤ -2.38) = 0.00866 = 0.0087 to 4d.p
b) The probability that a student will complete the exam in more than 60 minutes but less than 75 minutes. P(60 < x < 75)
We first standardize/normalize 60 and 75 minutes
For 60 minutes
z = (x - μ)/σ = (60 - 79)/8 = - 2.38
For 75 minutes
z = (x - μ)/σ = (75 - 79)/8 = - 0.50
To determine the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes.
P(60 < x < 75) = P(-2.38 < z < -0.50)
We'll use data from the normal probability table for these probabilities
P(60 < x < 75) = P(-2.38 < z < -0.50)
= P(z < -0.50) - P(z < -2.38)
= 0.30954 - 0.00866
= 0.30088 = 0.3009
c.
Probability of finishing in 90 minutes or less = P(x ≤ 90)
We first standardize/normalize 90 minutes
z = (x - μ)/σ = (90 - 79)/8 = 1.38
To determine the probability of completing the exam in 90 minutes or less.
P(x ≤ 90) = P(z ≤ 1.38)
P(x ≤ 90) = P(z ≤ 1.38) = 0.91621
Thus, 91.621% of the class will finish just in time.
About 54 students will finish in time.
Meaning about 6 students will be unable to finish the exam in the allowed time.
Find more information about probability here:
brainly.com/question/24756209
