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MATH 225N Week 5 Understanding Normal Distribution (QUESTIONS & ANSWERS) $15.99   Add to cart

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MATH 225N Week 5 Understanding Normal Distribution (QUESTIONS & ANSWERS)

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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...

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  • February 8, 2022
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  • 2021/2022
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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/

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