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The time needed to complete a final examination in a particular college course is normally distributed with a mean of 8...

The time needed to complete a final examination in a particular college course is normally distributed with a mean of 80 minutes and a standard deviation of 10 minutes. Answer the following questions:

A) What is the probability of completing the exam in ONE hour or less?

B) what is the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes?

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 compete the exam in the allotted time?

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Answer #1
Concepts and reason

Normal distribution:

Normal distribution is a continuous distribution of data that has the bell-shaped curve. The normally distributed random variable x has meanand standard deviation.

Also, the standard normal distribution represents a normal curve with mean 0 and standard deviation 1. Thus, the parameters involved in a normal distribution are mean and standard deviation.

Standardized z-score:

The standardized z-score represents the number of standard deviations the data point is away from the mean.

If the z-score takes positive value when it is above the mean

If the z-score takes negative value when it is below the mean

Fundamentals

LetX-N(u,0)
, then the standard z-score is found using the formula given below:

11-X

Where X denotes the individual raw score, denotes the population mean and denotes the population standard deviation.

The different probabilities are calculated as follows:

P(Z > a)=1-P(Z <a)

Plasz sb)= P(Z <b) - P(Z <a)

The Excel-MegaStat procedure for finding the probability value is follows:

1.In EXCEL, Select Add-Ins > MegaStat > Probability.

2.Choose Continuous probability distributions.

3.Select Normal distribution and select calculate P given z and enter z as ____.

4.Enter mean as ____ and standard deviation as ___.

5.Click Ok.

(A)

The probability of completing the exam in one hour or less is calculated as follows:

Given information that time for the final examination is normally distributed with mean 80 minutes and standard deviation 10 minutes. That is, 08 = 11
ando=10
.

Consider 1 hour as 60 minutes.

The probability is

P(X360) =r(zsX7
=P(2500180)
= P(Z <-2)

Instructions to find the probability value for normal distribution:

1.In EXCEL, Select Add-Ins > MegaStat > Probability.

2.Choose Continuous probability distributions.

3.Select Normal distribution and select calculate P given z and enter z as –2.

4.Enter mean as 0 and standard deviation as 1.

5.Click Ok.

Follow the above instructions to get the following output:

EN
z
3
To
-2.00
Normal distribution
P(lower)
.0228
P(upper)
9772
z
-2.00

The probability value is 0.9772. That is, P(Z <-2)=0.0228

Thus, the probability of completing final exam in one hour or less is 0.0228.

(B)

The probability that a student will complete the exam in more than 60 minutes but less than 75 minutes is calculated as follows:

P(60 <X<75) = P( 60-1 X-μ 75-4)
σ σ σ
( 60-80 75 - 80)
( 10 10)
= P(-2<Z<-0.5)
= P(Z<-0.5)- P(Z<-2)

From the standard normal table, P(Z <-2)=0.0228
and P(Z <-0.5)=0.3085
.

P(60< X < 75) = P(Z <-0.5) - P(Z<-2)
= 0.3085-0.0228
= 0.2857

(C)

The expected number of students unable to finish the exam in 90 minutes is calculated as follows:

From the given information, the total number of students is 60.

The probability of person unable to finish the exam in 90 minutes is calculated as:

P(Z >90)=1- P(Z <90)
=1-P(z<995)
=1- P(Z <1)

From the standard normal table, P(Z <1) = 0.8413
.

P(Z > 90)=1- P(Z <1)
=1-0.8413
= 0.1587

From 60 students, the expected number of students unable to complete the exam in 90 minutes is,

60x0.1587=9.52
-10

Ans: Part A

Thus, the probability of completing final exam in one hour or less is 0.0228.

> Verified correct

Gissel Barragan Sat, Oct 30, 2021 7:55 PM

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