Normal Distribution Question. Need help

In summary, the conversation discusses how to calculate the probability of a random variable following a normal distribution with a mean of 80 and a standard deviation of 24 being outside the range of 32 and 116. One approach is to use the formula P(a < X < b) = P(X < b) - P(X < a) and find the probabilities using standard normal tables. The final result is 1 - 0.0228 - 0.0668, but it is also possible to arrive at this answer by breaking down the problem into smaller pieces.
  • #1
helix999
32
0
[MENTOR note] Post moved from General Math forum hence no template.

Assume that a random variable follows a normal distribution with a mean of 80 and a standard deviation of 24. What percentage of this distribution is not between 32 and 116?
My approach is to calculate the Probability for (mean - 2*σ < X < mean + 1.5*σ), not sure how to solve this further.
 
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  • #3
helix999 said:
[MENTOR note] Post moved from General Math forum hence no template.

Assume that a random variable follows a normal distribution with a mean of 80 and a standard deviation of 24. What percentage of this distribution is not between 32 and 116?
My approach is to calculate the Probability for (mean - 2*σ < X < mean + 1.5*σ), not sure how to solve this further.

Use ##P(a < X < b) = P(X < b) - P(X < a)##, and find both ##P(X < a)## and ##P(X < b)## from standard normal tables.
 
  • #4
Ray Vickson said:
Use ##P(a < X < b) = P(X < b) - P(X < a)##, and find both ##P(X < a)## and ##P(X < b)## from standard normal tables.
The solution I have with me is:
Prob(mean - 2*σ < X < mean + 1.5*σ) = (0.5 - 0.0228) + (0.5 - 0.0668)
My question is how we got (0.5 - 0.0228) + (0.5 - 0.0668) ?
 
  • #5
Maybe start here: http://stattrek.com/m/probability-distributions/standard-normal.aspx
 
  • #6
Try to break the problem into smaller pieces. What probability do you believe 0.5 - 0.0228 is specifying? (i.e. Probability that X is in what range?).
 
  • #7
helix999 said:
The solution I have with me is:
Prob(mean - 2*σ < X < mean + 1.5*σ) = (0.5 - 0.0228) + (0.5 - 0.0668)
My question is how we got (0.5 - 0.0228) + (0.5 - 0.0668) ?
Try both your method and the one suggested by Ray. They should give the same result. Drawing a picture might help as well.
It might also be interesting to note that (0.5 - 0.0228) + (0.5 - 0.0668) = 1 - 0.0228 - 0.0668. Do you know why those two numbers (0.0228 and 0.0668) are significant in this problem?
 

Related to Normal Distribution Question. Need help

What is a normal distribution?

A normal distribution is a statistical distribution that is symmetrical and bell-shaped. It is characterized by a mean, median, and mode that are all equal, and the majority of the data falls within one standard deviation of the mean.

How is a normal distribution different from other distributions?

A normal distribution is different from other distributions because it has a specific shape and specific properties, such as the mean, median, and standard deviation being equal. Other distributions may have different shapes and properties.

What are the properties of a normal distribution?

The properties of a normal distribution include symmetry, a bell-shaped curve, a mean, median, and mode that are all equal, and the majority of the data falling within one standard deviation of the mean. It also follows the 68-95-99.7 rule, where 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.

How is a normal distribution used in statistics?

A normal distribution is used in statistics to analyze and interpret data, make predictions, and conduct hypothesis testing. It is also used to calculate probabilities and determine the likelihood of certain events occurring.

What are some real-life examples of a normal distribution?

Some real-life examples of a normal distribution include height and weight measurements, test scores, and IQ scores. Many natural phenomena, such as the distribution of rainfall or the size of animal populations, also follow a normal distribution.

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