Now that you understand how to identify a normal distribution, we can utilize R to perform calculations that are specific to this distribution type.

When a normal distribution has been identified, we can estimate the probability that an event takes place as it occurs between two values.

I am assuming that you have some understanding of normal distributions in addition to what was discussed in the prior entry.

When a normal distribution has been identified, we can estimate the probability that an event takes place as it occurs between two values.

I am assuming that you have some understanding of normal distributions in addition to what was discussed in the prior entry.

__Example:__You are currently employed as a statistician in a factory that produces flashlights. The senior statistician informs you that the premium brand of flashlights that the factory produces, have a battery life expectancy which is normally distributed. The mean for the battery life of this particular brand is 20 hours, with a standard deviation of 5 hours.

__What is the probability that a randomly selected flashlight from the production line will last between 20-25 hours?__

**pnorm(q=25, mean=20, sd=5, lower.tail=TRUE)**

*# Output = 0.8413447 #*

**0.8413447 - .50**

*# 0.3413447 or % 34.134 Probability #*

__If a flashlight's battery dies at exactly 8 hours after use, how many standard deviations away from the mean is this value?__

*# (x - mean) / standard deviation #*

**(8 - 20) / 5**

*# - 2.4 Standard Deviations #*

__What is the probability that a randomly selected flashlight from the production line will last between 18-24 hours?__

**pnorm(q=18, mean=20, sd=5, lower.tail=FALSE)**

*# Output = 0.6554217 #*

**0.6554217 - .50**

*# Output = 0.1554217 #*

**pnorm(q=24, mean=20, sd=5, lower.tail=TRUE)**

*# Output = 0.7881446 #*

**0.7881446 - .50**

*# Output = 0.2881446 #*

**0.2881446 + 0.1554217**

*# 0.4435663 or % 44.357 Probability #*

__What is the probability that a randomly selected flashlight from the production line will last between 22-26 hours?__

**pnorm(q=22, mean=20, sd=5, lower.tail=FALSE)**

*# Output = 0.3445783 #*

**.50 - 0.3445783**

*# Output = 0.1554217 #*

**pnorm(q=26, mean=20, sd=5, lower.tail=TRUE)**

*# Output = 0.8849303 #*

**0.8849303 - 0.1554217 - .50**

# 0.2295086 or % 22.950 Probability #

# 0.2295086 or % 22.950 Probability #

At the same factory, while eating lunch, the senior statistician appears again. During this encounter, he decides to test your statistical abilities by asking you a series of questions.

These questions are:

__Given a normal distribution with a mean of 55, what is the standard deviation if 45% of the values are above 70?__

**qnorm(.45, lower.tail=FALSE)**

# Output = 0.1256613 #

# Output = 0.1256613 #

**70 - 55**

*# Output = 15 #*

**15 / .1256613**

*# Standard Deviation = 119.3685 #*

__Given a normal distribution with a standard deviation of 15, what is the mean if 25% of the values are below 45?__

**qnorm(.25, lower.tail = FALSE)**

*# Output =*

*0.6744898 #*

**45 + (0.6744898 * 15)**

*# Output = 55.11735 #*

Given a normal distribution with 60% of the values above 100, and 90% of the values above 80, what are the mean and the standard deviation?

**qnorm(.60, lower.tail=TRUE)**

*# Output =*0.2533471 #

**qnorm(.90, lower.tail=TRUE)**

*# Output =*1.281552 #

# (100 - Mean)/Standard Deviation = 0.2533471 #

# (80 - Mean)/Standard Deviation = 1.281552 #

# 100 - Mean = 0.2533471 * Standard Deviation #

# 80 - Mean = 1.281552 * Standard Deviation #

Which can be worked out, algebraically, to solve for both mean and standard deviation.

That is all of this entry, which closes out the 50th article that I have written for this blog. Two things to remember about normal distributions: there is no perfect test for normality, and there is no way to provide a probability for a single event occurring within a continuous normal distribution. All that we can find, is the probability surrounding an event's parameters.

Stay tuned until next time, Data Heads.

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