According to the Empirical Rule, 68% of All Newborn Babies in the United States Weigh Between and .

Objective

Here you will larn how to use the Empirical Dominion to estimate the probability of an event.

If the price per pound of USDA Choice Beefiness is normally distributed with a mean of $4.85/lb and a standard divergence of $0.35/lb, what is the estimated probability that a randomly chosen sample (from a randomly chosen market) will be between $five.twenty and $5.55 per pound?

Scout This: Empirical Rule

Guidance

This reading on the Empirical Rule is an extension of the previous reading "Agreement the Normal Distribution." In the prior reading, the goal was to develop an intuition of the interaction betwixt decreased probability and increased distance from the mean. In this reading, nosotros volition practice applying the Empirical Rule to estimate the specific probability of occurrence of a sample based on the range of the sample, measured in standard deviations.

The graphic beneath is a representation of the Empirical Rule:

The graphic is a rather concise summary of the vital statistics of a Normal Distribution. Note how the graph resembles a bell? At present you know why the normal distribution is also called a " bell bend."

• 50% of the information is above, and l% below, the mean of the data
• Approximately 68% of the data occurs inside 1 SD of the hateful
• Approximately 95% occurs inside ii SD'south of the mean
• Approximately 99.7% of the data occurs within 3 SDs of the hateful

It is due to the probabilities associated with one, 2, and three SDs that the Empirical Dominion is too known as the 68−95−99.seven dominion.

Instance 1

If the diameter of a basketball is normally distributed, with a mean (µ) of 9″, and a standard deviation (σ) of 0.five″, what is the probability that a randomly chosen basketball will accept a bore between 9.5″ and x.5″?

Solution

Since the σ = 0.5″ and the µ = 9″, we are evaluating the probability that a randomly called ball volition have a diameter between 1 and 3 standard deviations above the hateful. The graphic below shows the portion of the normal distribution included between 1 and iii SDs:

The percent of the information spanning the 2nd and 3rd SDs is xiii.5% + 2.35% = fifteen.85%

The probability that a randomly called basketball game will have a diameter betwixt 9.5 and 10.v inches is 15.85%.

Example 2

If the depth of the snow in my yard is normally distributed, with µ = two.5″ and σ = .25″, what is the probability that a randomly called location will have a snowfall depth between two.25 and 2.75 inches?

Solution

2.25 inches is µ − 1σ, and two.75 inches is µ + 1σ, so the area encompassed approximately represents 34% + 34% = 68%.

The probability that a randomly chosen location will have a depth betwixt 2.25 and 2.75 inches is 68%.

Instance 3

If the height of women in the United states is normally distributed with µ = v′ eight″ and σ = 1.5″, what is the probability that a randomly called woman in the United States is shorter than v′ 5″?

Solution

This ane is slightly different, since we aren't looking for the probability of a limited range of values. We want to evaluate the probability of a value occurring anywhere below 5′ 5″. Since the domain of a normal distribution is infinite, nosotros tin can't actually country the probability of the portion of the distribution on "that end" because it has no "end"! What we need to exercise is add upward the probabilities that nosotros exercise know and subtract them from 100% to get the residue.

Here is that normal distribution graphic once again, with the tiptop data inserted:

Recall that a normal distribution always has 50% of the data on each side of the hateful. That indicates that 50% of US females are taller than five′ 8″, and gives us a solid starting point to calculate from. There is some other 34% between 5′ half dozen.5″ and v′ eight″ and a concluding 13.five% betwixt v′ 5″ and 5′ 6.5″. Ultimately that totals: l% + 34% + 13.5% = 97.5%. Since 97.5% of Usa females are 5′ 5″ or taller, that leaves 2.5% that are less than 5′ 5″ tall.

Intro Problem Revisited

If the toll per pound of USDA Pick Beefiness is normally distributed with a mean of $iv.85/lb and a standard departure of $0.35/lb, what is the estimated probability that a randomly chosen sample (from a randomly chosen market) will be betwixt $5.20 and $5.55 per pound?

$five.twenty is µ + 1σ, and $5.55 is µ + 2σ, so the probability of a value occurring in that range is approximately 13.5%.

Vocabulary

Normal distribution: a common, merely specific, distribution of data with a set of characteristics detailed in the lesson in a higher place.

Empirical Rule: a name for the way in which the normal distribution divides data by standard deviations: 68% within 1 SD, 95% within 2 SDs and 99.7 within 3 SDs of the mean

68-95-99.7 rule: another name for the Empirical Rule

Bell bend: the shape of a normal distribution

Guided Practice

1. A normally distributed information set has µ = 10 and σ = two.5, what is the probability of randomly selecting a value greater than 17.5 from the set?
2. A normally distributed data set has µ = .05 and σ = .01, what is the probability of randomly choosing a value between .05 and .07 from the set?
3. A normally distributed data fix has µ = 514 and an unknown standard departure, what is the probability that a randomly selected value will be less than 514?

Solutions

1. If µ = x and σ = 2.5, then 17.5 = µ + 3σ. Since we are looking for all data above that betoken, we need to decrease the probability that a value volition occur below that value from 100%: The probability that a value will be less than x is 50%, since 10 is the mean. There is another 34% between ten and 12.v, some other 13.5% between 12.5 and 15, and a final 2.35% betwixt 15 and 17.five. 100% −50% −34% −13.five% −two.35% = 0.15% probability of a value greater than 17.five
2. 0.05 is the mean, and 0.07 is two standard deviations above the hateful, then the probability of a value in that range is 34% + 13.v% = 47.5%
3. 514 is the mean, so the probability of a value less than that is fifty%.

Practice Questions

Assume all distributions to exist normal or approximately normal, and summate percentages using the 68−95−99.7 rule.

1. Given hateful 63 and standard difference of 168, observe the guess percentage of the distribution that lies betwixt −105 and 567.
2. Approximately what percent of a normal distribution is between two standard deviations and 3 standard deviations from the mean?
3. Given standard difference of 74 and hateful of 124, approximately what percentage of the values are greater than 198?
4. Given σ = 39 and µ = 101, approximately what percentage of the values are less than 23?
5. Given mean 92 and standard deviation 189, find the approximate pct of the distribution that lies between −286 and 470.
6. Approximately what percent of a normal distribution lies betwixt µ + 1σ and µ + 2σ?
7. Given standard deviation of 113 and mean 81, approximately what pct of the values are less than −145?
8. Given mean 23 and standard deviation 157, find the approximate percentage of the distribution that lies between 23 and 337.
9. Given σ = iii and µ = 84, approximately what per centum of the values are greater than 90?
10. Approximately what percent of a normal distribution is between µ and µ+1σ?
11. Given mean 118 and standard deviation 145, find the guess percentage of the distribution that lies betwixt −27 and 118.
12. Given standard departure of 81 and mean 67, approximately what pct of values are greater than 310?
13. Approximately what per centum of a normal distribution is less than 2 standard deviations from the hateful?
14. Given µ + 1σ = 247 and µ + 2σ = 428, find the approximate percentage of the distribution that lies between 66 and 428.
15. Given µ − 1σ = −131 and µ + 1σ = 233, approximately what percentage of the values are greater than −495?

According to the Empirical Rule, 68% of All Newborn Babies in the United States Weigh Between and .

Source: https://courses.lumenlearning.com/atd-austincc-mathlibarts/chapter/the-empirical-rule/

Share :