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Normal Distribution

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Statistics

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Homework

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Q1.
For a normal distribuon curve with a mean of 19 and a standard deviaon of 6, which range of the variable denes an area under the curve corresponding to a probability
of approximately 68%?
Ans
Normal Distribuon
Mean ( u ) =19
Standard Deviaon ( sd )=6
Normal Distribuon = Z= X- u / sd ~ N(0,1)
To nd P(a < = Z < = b) = F(b) - F(a)
P(X < 13) = (13-19)/6
= -6/6 = -1
= P ( Z <-1) From Standard Normal Table
= 0.15866
P(X < 25) = (25-19)/6
= 6/6 = 1
= P ( Z <1) From Standard Normal Table
= 0.84134
P(13 < X < 25) = 0.84134-0.15866 = 0.6827 ~ 68%
Q2.
What is the area under the standard normal distribuon curve between z = 1.50 and z = 2.50?
Ans
To nd P(a ≤ Z ≤ b) = F(b) - F(a)
P(X < 1.5) = X- u / sd = P( X < Z ) [ We Have Value of Z = 1.5 ]
= P ( X < 1.5 ) From Standard Normal Table
= 0.9332
P(X < 2.5) = X- u / sd = P( X < Z ) [ We Have Value of Z = 2.5 ]
= P ( X < 2.5 ) From Standard Normal Table
= 0.9938
And P(1.5 < X < 2.5) = 0.0606
Q3.
The average height of @owering cherry trees in a certain nursery is 9.5 feet. If the heights are normally distributed with a standard deviaon of 1.3 feet, nd the probability
that a tree is less than 11.5 feet tall?
Ans
Normal Distribuon
Mean ( u ) =9.5
Standard Deviaon ( sd )=1.3
Normal Distribuon = Z= X- u / sd ~ N(0,1)
P(X < 11.5) = (11.5-9.5)/1.3
= 2/1.3= 1.5385
= P ( Z <1.5385) From Standard Normal Table
= 0.94
Q4.
If a normally distributed group of test scores have a mean of 70 and a standard deviaon of 12, nd the percentage of scores that will fall below 50?
Ans
Normal Distribuon
Mean ( u ) =70
Standard Deviaon ( sd )=12
Normal Distribuon = Z= X- u / sd ~ N(0,1)
P(X < 50) = (50-70)/12
= -20/12= -1.6667
= P ( Z <-1.6667) From Standard Normal Table
= 0.0475~ 4.75%
Q5.
For a normal distribuon with a mean of 13 and a standard deviaon of 6, the value –2 has a z value of
Ans
P(X < -2) = (-2-13)/6
= -15/6= -2.5
Q6.
What is the area under the standard normal distribuon curve between z = 0 and z = - 2.16 ?

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Ans
To nd P(a ≤ Z ≤ b) = F(b) - F(a)
P(X < 0) = X- u / sd = P( X < Z ) [ We Have Value of Z = 0 ]
= P ( X < 0 ) From Standard Normal Table
= 0.5
P(X < -2.16) = X- u / sd = P( X < Z ) [ We Have Value of Z = -2.16 ]
= P ( X < -2.16 ) From Standard Normal Table
= 0.0154
P(0 < X < -2.16) = -0.4846
By symmetry there is a point to the right of the mean where z = +0.4846
Q7.
In a standard normal distribuon, what z value corresponds to 17% of the data between the mean and the z value?
Ans
Draw the picture and you will see there is a leE-tail of 50-17 = 33%.
From Standard Normal Table Z(0.33) = -0.4399
By symmetry there is a point to the right of the mean where z = +0.4399
Q8.
Give the term for the number of standard deviaons that a parcular X value is away from the mean
Ans
It is Z-value
Q9.
In order to be accepted into a top university, applicants must score within the top 5% on the SAT exam. Given that the test has a mean of 1000 and a standard deviaon of
200, what is the lowest possible score a student needs to qualify for acceptance into the university?
Ans
P ( Z > x ) = 0.95
Value of z to the cumulave probability of 0.95 from normal table is 1.64
P( x-u/ (s.d/sqrt(n) ) > x - 1000/ (200/sqrt(1) ) = 0.95
That is, ( x - 1000/ (200/sqrt(1) ) = 1.64
--> x = 1.64 * 200/sqrt(1) + 1000 = 1328.9707 ~ 1330
Q10.
At a large department store, the average number of years of employment for a cashier was 5.7 with a standard deviaon of 3.8 years. For a sample of 50 employees, what is
the probability that the sample mean is greater than 7 years?
Ans
Normal Distribuon
Mean ( u ) =5.7
Standard Deviaon ( sd )=3.8
Number ( n ) = 50
Normal Distribuon = Z= X- u / (sd/Sqrt(n) ~ N(0,1)
P(X > 7) = (7-5.7)/3.8/ Sqrt ( 50 )
= 1.3/0.537= 2.419
= P ( Z >2.419) From Standard Normal Table
= 0.0078 ~ 0.78%

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Q1.For a normal distribution curve with a mean of 19 and a standard deviation of 6, which range of the variable defines an area under the curve corresponding to a probability of approximately 68%?AnsNormal DistributionMean ( u ) =19Standard Deviation ( sd )=6Normal Distribution = Z= X- u / sd ~ N(0,1)To find P(a < = Z < = b) = F(b) - F(a)P(X < 13) = (13-19)/6 = -6/6 = -1 = P ( Z <-1) From Standard Normal Table = 0.15866P(X < 25) = (25-19)/6 = 6/6 = 1 = P ( Z <1) From Standard Normal Table = 0.84134P(13 < X < 25) = 0.84134-0.15866 = 0.6827 ~ 68%Q2.What is the ...
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