Answer:
a. The probability of completing the exam in one hour or less is 0.0783
b. The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is 0.3555
c. The number of students will be unable to complete the exam in the allotted time is 8
Step-by-step explanation:
a. According to the given we have the following:
The time for completing the final exam in a particular college is distributed normally with mean (μ) is 77 minutes and standard deviation (σ) is 12 minutes
Hence, For X = 60, the Z- scores is obtained as follows:
Z= X−μ
/σ
Z=60−77
/12
Z=−1.4167
Using the standard normal table, the probability P(Z≤−1.4167) is approximately 0.0783.
P(Z≤−1.4167)=0.0783
Therefore, The probability of completing the exam in one hour or less is 0.0783.
b. In this case For X = 75, the Z- scores is obtained as follows:
Z= X−μ
/σ
Z=75−77
/12
Z=−0.1667
Using the standard normal table, the probability P(Z≤−0.1667) is approximately 0.4338.
Therefore, The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is obtained as follows:
P(60<X<75)=P(Z≤−0.1667)−P(Z≤−1.4167)
=0.4338−0.0783
=0.3555
Therefore, The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is 0.3555
c. In order to compute how many students you expect will be unable to complete the exam in the allotted time we have to first compute the Z−score of the critical value (X=90) as follows:
Z= X−μ
/σ
Z=90−77
/12
Z=1.0833
UsING the standard normal table, the probability P(Z≤1.0833) is approximately 0.8599.
Therefore P(Z>1.0833)=1−P(Z≤1.0833)
=1−0.8599
=0.1401
Therefore, The number of students will be unable to complete the exam in the allotted time is= 60×0.1401=8.406
The number of students will be unable to complete the exam in the allotted time is 8
≈ -1.31
