Dr. B. F. Goodrich is trying to see how well her statistics students are doing w

Question

Dr. B. F. Goodrich is trying to see how well her statistics students are doing with mastering the concept of standard deviation. She believes that the better students understand the concept the faster they can work problems, so she times her students individually to see how long they take to compute a standard deviation. The raw scores below present the number of minutes the students in her class took to solve the problem. (Hint for this problem you have to calculate the mean and standard deviation) 9.0 5.0 6.0 7.0 9.0 4.0 6.0 7.0 10.0 Find the z scores for each of the raw scores in this sample. Use the unit normal table to find the following: p (x> 5.5)= p(x

Explanation / Answer

Solution:

For the given data, we have

Mean = 7

SD = 2

Now, we have to find Z-scores for each of the raw score in the given sample.

Z scores are given as below:

Z = (X – mean) / SD

X

Z

9

(9 – 7)/2 = 1

5

(5 – 7)/2 = -1

6

(6 – 7)/2 = -0.5

7

(7 – 7)/2 = 0

9

(9 – 7)/2 = 1

4

(4 – 7)/2 = -1.5

6

(6 – 7)/2 = -0.5

7

(7 – 7)/2 = 0

10

(10 – 7)/2 = 1.5

Now, we have to find P(X>5.5)

P(X>5.5) = 1 – P(X<5.5)

Z = (X – mean) / SD

Z = (5.5 – 7)/2 = -0.75

P(Z< -0.75) = P(X<5.5) = 0.226627

P(X>5.5) = 1 – P(X<5.5) = 1 - 0.226627 = 0.773373

Required probability = 0.773373

Now, we have to find P(X<6.0)

Z = (6 – 7) / 2 = 0.5

P(Z<0.5) = P(X<6.0) = 0.691462

Required probability = 0.691462

Now, we have to find P(6.9<X<10)

P(6.9<X<10) = P(X<10) – P(X<6.9)

For X<10

Z = (10 – 7) / 2 = 3/2 = 1.5

P(Z<1.5) = P(X<10) = 0.933193

For X<6.9

Z = (6.9 – 7) / 2 = -0.05

P(Z< -0.05) = P(X<10) = 0.480061

P(6.9<X<10) = P(X<10) – P(X<6.9)

P(6.9<X<10) = 0.933193 - 0.480061 = 0.453132

Required probability = 0.453132

Now, we have to find % of scores greater than 5.5

P(X>5.5) = 1 – P(X<5.5)

Z = (5.5 – 7) / 2 = -0.75

P(Z< -0.75) = P(X<5.5) = 0.226627

P(X>5.5) = 1 – P(X<5.5) = 1 - 0.226627 = 0.773373

Required probability = 0.773373

Required percentage = 77.34%

Now, we have to find score for middle 95%

Remaining area = 1 – 0.95 = 0.05

Area at left = 0.05/2 = 0.025

Area at right = 0.05/2 = 0.025

Z for left = -1.95996

Z for right = 1.95996

X = Mean + Z*SD

XL = 7 -1.95996*2 = 3.08008

XU = 7 + 1.95996*2 =10.91992

Required scores = 3.08008 and 10.91992

Now, we have to find score for extreme 30%

That is, we have to find score for lower 70%

Z for lower 70% or extreme 30% is 0.524401

X = Mean + Z*SD

X = 7 + 0.524401*2 = 8.048801

Required score = 8.048801

Now, we have to find score for upper 5%

Z for upper 5% = 1.644854

X = Mean + Z*SD

X = 7 + 1.644854*2 = 10.28971

Required score = 10.28971

Raw scores with mean = 100 and SD = 15 are given as below:

X = Mean + Z*SD

Z

Raw score X

1

100 + 1*15 = 115

-1

100 – 1*15 = 85

-0.5

100 – 0.5*15 = 92.5

0

100 + 0*15 = 100

1

100 + 1*15 = 115

-1.5

100 – 1.5*15 = 77.5

-0.5

100 – 0.5*15 = 92.5

0

100 + 0*15 = 100

1.5

100 + 1.5*15 = 122.5

X

Z

9

(9 – 7)/2 = 1

5

(5 – 7)/2 = -1

6

(6 – 7)/2 = -0.5

7

(7 – 7)/2 = 0

9

(9 – 7)/2 = 1

4

(4 – 7)/2 = -1.5

6

(6 – 7)/2 = -0.5

7

(7 – 7)/2 = 0

10

(10 – 7)/2 = 1.5

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