Question

1. Certain standardized math exams have a mean of 100 and a standard deviation of 60. Of a sample of 36 students who take this exam, what percent could you expect to score between 100 and 120?

A)50

B)47.5

C)97.5

D)49.85

2.Certain standardized math exams have a mean of 100 and a standard deviation of 60. Of a sample of 36 students who take this exam, what percent could you expect to score above 120?

A)2.5

B)2.35

C)97.5

D)13.5

3. Certain standardized math exams have a mean of 100 and a standard deviation of 60. Of a sample of 36 students who take this exam, what percent could you expect to score above 70?

 A) 99.85 B) 95 C) 97.5 D) 0.15

4.Certain standardized math exams have a mean of 100 and a standard deviation of 60. Of a sample of 36 students who take this exam, what percent could you expect to score between 80 and 110?

A)84

B)81.5

C)83.85

D)85

5.Certain standardized math exams had a mean of 120 and a standard deviation of 20. Of students who take this exam, what percent could you expect to score below 100?

6. A mean could be a

A) Statistics only

B) Parameter only

C) Statistics and parameter

Solution :

Given that ,

mean = = 100

standard deviation = = 60

1)

n = 36  = 100  = / n = 60 / 36 = 60 / 6 = 10

P(100 < < 120) = P((100 - 100) / 10<( -  ) /  < (120 - 100) / 10))

= P(0 < Z < 2)

= P(Z < 2) - P(Z < 0)

= 0.9750 - 0.5

= 0.475

Percent = 47.5

2)

P( > 120) = 1 - P( < 120)

= 1 - P(( -  ) /  < (120 - 100) / 10)

= 1 - P(z < 2)

= 1 - 0.975

= 0.025

Percent = 2.5%

3)

P( < 70) = P(( -  ) /  < (70 - 100) / 10)

= P(z < -3)

= 0.9985

Percent = 99.85

4)

P(80 < < 110) = P((80 - 100) / 10<( -  ) /  < (110 - 100) / 10))

= P(-2 < Z < 1)

= P(Z < 1) - P(Z < -2)

= 0.84 - 0.025

= 0.815

Percent = 81.5

5)

P( < 100) = P(( -  ) /  < (100 - 100) / 10)

= P(z < 0)

= 0.50

Percent = 50%

6)

A mean could be a Statistics and parameter .

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