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AFor exercises 1 and 2, a set of data has a normal distribution with a mean of 120 and a standard deviation of 10.

  1. Find the interval about the mean within which 90% of the data lie.
A94.2-145.8
B103.5-136.5
C113.68-126.32
D100.4-139.6

  1. Find the probability that a value selected at random from this data is between 100 and 140.
A99.9%
B90%
C99%
D95.5%

BIn a random sample of 1000 exams, the average score was 500 points with a standard deviation of 80 points.

  1. Find the interval about the sample mean that has a 1% level of confidence.
A499-501
B479-520
C495-505
D493-507

  1. Find the interval about the sample mean such that the probability is 0.90 that the mean number lies within the interval.
A499-501
B495-505
C496-504
D368-632

  1. Find the probability that the mean score of the population will be less than five points from the mean score of the sample.
A95.5%
B38.3%
C98.8%
D62.5%

  1. Find the probability that the true mean is between 495 and 500.
A95%
B47.75%
C95.5%
D68.3%

Verified Answer :

nobillionaireNobley

A1.) the interval about the mean within which 90% of the data lie = 120 + or - 1.645(10) = 120 + or - 16.45 = 120 - 16.45 to 120 + 16.45 = 103.55 to 136.45

2.) P(100 < X < 140) = P(X < 140) - P(X < 100) = P(z < (140 - 120)/10) - P(z < (100 - 120)/10) = P(z < 2) - P(z < -2) = P(z < 2) - [1 - P(z < 2)] = 2P(z < 2) - 1 = 2(0.97725) - 1 = 1.9545 - 1 = 0.9545 = 95.5%

B1.) the interval about the sample mean that has a 1% level of confidence is 500 + or - 2.575(80/√1000) = 500 + or - 7 = 500 - 7 to 500 + 7 = 493 to 507

2.) 2P(z < a) - 1 = 0.90
2P(z < a) = 1.90
P(z < a) = 0.95
a = 1.645
(b - 500)/(80/√1000) = 1.645
b - 500 = 4
b = 500 + 4 = 504
The interval about the sample mean such that the probability is 0.90 that the mean number lies within the interval is 496 - 504.