MATH 225N Week 5 Assignment: Applications of the Normal Distribution
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Course
MATH 225N (MATH225N)
Institution
Chamberlain College Of Nursing
Question
Sugar canes have lengths, X , that are normally distributed with mean 365.45
centimeters and standard deviation 4.9 centimeters. What is the probability of the length
of a randomly selected cane being between 360 and 370 centimeters?
Round your answer to four decimal places.
The m...
week 5 assignment applications of the normal distribution
math 225n week 5 assignment applicat
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Week 5 Assignment: Applications of the
Normal Distribution - Excel
Question
Sugar canes have lengths, X , that are normally distributed with mean 365.45
centimeters and standard deviation 4.9 centimeters. What is the probability of the length
of a randomly selected cane being between 360 and 370 centimeters?
Round your answer to four decimal places.
The mean is μ=365.45 , and the standard deviation is σ=4.9 . As the probability
between two values is to be calculated, subtract the probability of the lower value from the
higher value. In this case, you have to use the NORMDIST function twice.
1. Open Excel and click on any empty cell. Click Insert function, fx .
2. Search for NORMDIST in the search for a function dialog box and click GO.
3. Make sure NORMDIST is on top in select a function. Then click OK.
4. In the function arguments of NORMDIST function, enter 370 for X , 365.45 for
Mean, 4.9 for Standard_dev, and TRUE for Cumulative, all for the higher value of X .
Thus, the answer, rounded to four decimal places, is 0.8234 .
5. Click on any other empty cell. Click Insert function, fx .
6. Search for NORMDIST in the search for a function dialog box and click GO.
7. Make sure NORMDIST is on top in select a function. Then click OK.
8. In the function arguments of NORMDIST function, enter 360 for X , 365.45 for
Mean, 4.9 for Standard_dev, and TRUE for Cumulative, all for the lower value of X .
Thus, the answer, rounded to four decimal places, is 0.1330 .
,Now subtract, 0.8234−0.1330=0.6904 . Thus, the probability of the length of a
randomly selected cane being between 360 and 370 centimeters is 0.6904 .
Question
The number of miles a motorcycle, X, will travel on one gallon of gasoline is
modeled by a normal distribution with mean 44 and standard deviation 5. If
Mike starts a journey with one gallon of gasoline in the motorcycle, find the
probability that, without refueling, he can travel more than 50 miles.
Round your answer to four decimal places.
The mean is μ=44, and the standard deviation is σ=5. The probability that
Mike can travel, without refueling, more than 50 miles is shown below.
A normal curve is over a horizontal axis and is divided into 3 regions. Vertical
line segments extend from the horizontal axis to the curve at the mean,
44 and at 50. The right region is shaded.
First find the probability to the left of 50 and subtract from 1.
1. Open Excel and click on any empty cell. Click Insert function, fx.
, 2. Search for NORMDIST in the search for a function dialog box and click
GO.
3. Make sure NORMDIST is on top in select a function. Then click OK.
4. In the function arguments of NORMDIST function, enter 50 for X, 44 for
Mean, 5 for Standard_dev, and TRUE for Cumulative.
This probability, rounded to four decimal places, is 0.8849. Now
subtract, 1−0.8849=0.1151. Thus, the desired probability is
P(X>50)=0.1151.
Question
A worn, poorly set-up machine is observed to produce components whose
length X follows a normal distribution with mean 14 centimeters and
variance 9. Calculate the probability that a component is at least 12
centimeters long.
Round your answer to four decimal places.
The mean is μ=14, and the standard deviation is σ=9–√=3.
1. Open Excel and click on any empty cell. Click Insert function, fx.
2. Search for NORMDIST in the search for a function dialog box and click
GO.
3. Make sure NORMDIST is on top in select a function. Then click OK.
4. In the function arguments of NORMDIST function, enter 12 for X, 14 for
Mean, 3 for Standard_dev, and TRUE for Cumulative. The probability,
rounded to four decimal places, is P(X<12)≈0.2525.
P(X≥12), so subtract from 1 to get
The desired probability is
P(X≥12)=1−0.2525=0.7475.
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