Understanding Normal Distribution: week #5
1. Lexie averages 149 points per bowling game with a standard deviation of 14 points.
Suppose Lexie's points per bowling game are normally distributed. Let X= the number of
points per bowling game. Then X∼N(149,14).
Suppose Lexie scores 186 points in the game on Tuesday. The z-score
when x = 186 is 2.643 - no response given. The mean is 149 - no
response given.
This z-score tells you that x = 186 is 2.643- no response given standard
deviations to the right of the mean.
The z-score can be found using this formula:
z=x−μσ=186−149/14=3714≈2.643
The z-score tells you how many standard deviations the value x is above (to
the right of) or below (to the left of) the mean, μ. So, scoring 186 points is
2.643 standard deviations away from the mean. A positive value of z
means that that the value is above (or to the right of) the mean, which was
given in the problem: μ=149 points in the game.
2. Suppose X∼N(18,2), and x=22. Find and interpret the z-score of the
standardized normal random variable.
This study source was downloaded by 100000770861734 from CourseHero.com on 02-08-2022 11:47:48 GMT -06:00
https://www.coursehero.com/file/58202586/Understanding-Normal-Distribution-week-5docx/
, 2
The z-score when x=22 is . The mean
18
is .
2
This z-score tells you that x=22 is standard
deviations to the right of the mean.
3. Suppose X∼N(12.5,1.5), and x=11. Find and interpret the z-score of the
standardized normal random variable.
X is a normally distributed random variable with μ=12.5 (mean) and σ=1.5
(standard deviation). To calculate the z-score,
z=x−μσ=11−12.51.5=−1.51.5=−1
This means that x=11 is one standard deviation (1σ) below or to the left of the
mean. This makes sense because the standard deviation is 1.5. So, one standard
deviation would be (1)(1.5)=1.5, which is the distance between the mean
(μ=12.5) and the value of x (11).
This study source was downloaded by 100000770861734 from CourseHero.com on 02-08-2022 11:47:48 GMT -06:00
https://www.coursehero.com/file/58202586/Understanding-Normal-Distribution-week-5docx/
1. Lexie averages 149 points per bowling game with a standard deviation of 14 points.
Suppose Lexie's points per bowling game are normally distributed. Let X= the number of
points per bowling game. Then X∼N(149,14).
Suppose Lexie scores 186 points in the game on Tuesday. The z-score
when x = 186 is 2.643 - no response given. The mean is 149 - no
response given.
This z-score tells you that x = 186 is 2.643- no response given standard
deviations to the right of the mean.
The z-score can be found using this formula:
z=x−μσ=186−149/14=3714≈2.643
The z-score tells you how many standard deviations the value x is above (to
the right of) or below (to the left of) the mean, μ. So, scoring 186 points is
2.643 standard deviations away from the mean. A positive value of z
means that that the value is above (or to the right of) the mean, which was
given in the problem: μ=149 points in the game.
2. Suppose X∼N(18,2), and x=22. Find and interpret the z-score of the
standardized normal random variable.
This study source was downloaded by 100000770861734 from CourseHero.com on 02-08-2022 11:47:48 GMT -06:00
https://www.coursehero.com/file/58202586/Understanding-Normal-Distribution-week-5docx/
, 2
The z-score when x=22 is . The mean
18
is .
2
This z-score tells you that x=22 is standard
deviations to the right of the mean.
3. Suppose X∼N(12.5,1.5), and x=11. Find and interpret the z-score of the
standardized normal random variable.
X is a normally distributed random variable with μ=12.5 (mean) and σ=1.5
(standard deviation). To calculate the z-score,
z=x−μσ=11−12.51.5=−1.51.5=−1
This means that x=11 is one standard deviation (1σ) below or to the left of the
mean. This makes sense because the standard deviation is 1.5. So, one standard
deviation would be (1)(1.5)=1.5, which is the distance between the mean
(μ=12.5) and the value of x (11).
This study source was downloaded by 100000770861734 from CourseHero.com on 02-08-2022 11:47:48 GMT -06:00
https://www.coursehero.com/file/58202586/Understanding-Normal-Distribution-week-5docx/
