Z-Score
Empirical Rule
Z-Score Table
Probability
Surprise Me!
100
What Z-score is equivalent to being 1 standard deviation BELOW the mean?

Z = -1.0

100

Based on the empirical rule, what % of the data do you expect to fall within 1 standard deviation of the mean?

68%

100

Find Pr(Z<0)

0.50
100

Suppose the length of Python snakes follows a normal distribution with a mean of 100 inches and standard deviation of 10 inches. 

What is the probability you would find a Python less than 80 inches?

Z= (80 - 100)/10  =-2.0

Pr (Z<-2.0) = 0.02275 

or 0.025 if used Empirical rule

100

Find Pr(Z<0.42)

0.66276

200

What does a Z-score of 1.5 mean?

The value you observed is 1.5 standard deviations above the mean. 
200

Using a normal distribution, what percentage of the data fall more than 2 standard deviations BELOW the mean?

2.5%

200

Find Pr(Z<1.0)

0.8413

200

Suppose the length of Python snakes follows a normal distribution with a mean of 100 inches and standard deviation of 10 inches. 

What is the probability you would find a Python more than 110 inches?

Z=(110-100)/10 = 1.0

Pr (Z>1.0) = 1-0.8413 = 0.1587

Or 0.16 (16%) if using Empirical Rule

200

In a normal distribution, what proportion of observations lie above the mean?

50%

300
Test score follow a normal distribution with a  mean of 80 and standard deviation of 6. 


Noah scores a 91 on the test. Calculate the Z-score corresponding to Noah's test grade. 

Z = 1.83
300

Using a normal distribution, what percentage of the data fall 1 standard deviation ABOVE the mean?

16%

300

Find Pr(Z>-0.65)

0.74215

300

Suppose the length of Python snakes follows a normal distribution with a mean of 100 inches and standard deviation of 10 inches. 

What is the probability you would find a Python less than 85 inches?

Z=(85-100)/10 = -1.50

Pr(Z<-1.50)=0.06681

300

Suppose the average cost of a gallon of gas follows a normal distribution with a mean of $3.50 and a standard deviation of $0.42. 

What is the probability that a gallon of gas would cost $2.99 or less?

Z=(2.99 - 3.50)/0.42 = -1.21


Pr (Z<-1.21)=0.11314


400

The average cost of a speeding ticket follows a normal distribution with a mean of $175.00 and a standard deviation of $45. 


In Georgia, the average cost of a speeding ticket is $136. Find the Z-score associated with the average cost of a speeding ticket in Georgia. 

Z=-0.87

400

Suppose the number of calories on a menu item at Cheesecake Factory follows a normal distribution with a mean of 1500 calories and a standard deviation of 500. 

Approximately 68% of all menu item's calorie count falls between what two values?

1000 and 2000 calories

400

Find Pr(Z<-0.17)

0.43215

400

Suppose the length of Python snakes follows a normal distribution with a mean of 100 inches and standard deviation of 10 inches. 

What is the probability you would find a Python more than 97 inches?

Z=(97-100)/10 = -0.3

Pr(Z>-0.3)=1-0.38209 = 0.61791

400

Suppose the average lifespan of a Bearded Dragon is 12.2 years with a standard deviation of 5.2 years. 


What is the probability that Thor will live for at least 15 years?

Z=(15-12.2)/5.2 = 0.538

Pr(Z>0.54)=1-0.70540 = 0.29460

500

Dogs live on average 12 years with a SD of 3 years. 

Cats live on average 14 years with a SD of 4 years. 

What value is considered more "extreme", a cat that lives for 6 years or a dog that lives for 18 years?

Z-score (dog) =2.0

Z-score (cat) = -2.0


They are equivalent. 

500
The average price of a prom dress follows a normal distribution with a mean of $600 and a standard deviation of $200. What percentage of prom dresses cost LESS than $400?

16%

500

Find Pr(Z>1.64)

0.0505

500

Suppose the average college basketball arena holds 41,000 people with a standard deviation of 1,500. 


What percentage of stadiums could hold more than 43,000 people?

Z=(43,000-41,000)/(1,500) = 1.33

Pr(Z>1.33)=1-0.90824=0.09176

500

Suppose the average lifespan of a Bearded Dragon is 12.2 years with a standard deviation of 5.2 years. 


What is the probability that Thor will live less than 8 years?

Z=(8-12.2)/5.2 = -0.808

Pr(Z<-0.81) = 0.20897

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