Show all work please
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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