Given a test that is normally distributed with a mean of 100 and a standard deviation of 12, find: (a) the probability that a single score drawn at random will be greater than 110 (b) the probability that a sample of 25 scores will have a mean greater than 105 (c) the probability that a sample of 64 scores will have a mean greater than 105 (d) the probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

1 answer
Question:

Given a test that is normally distributed with a mean of 100 and a standard deviation of 12, find:
(a) the probability that a single score drawn at random will be greater than 110
(b) the probability that a sample of 25 scores will have a mean greater than 105
(c) the probability that a sample of 64 scores will have a mean greater than 105
(d) the probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

Answers

Answer:(a)  0.2033(b)  0.0188(c)  0.0004(d)  0.095Step-by-step explanation:(a) the probability that a score at random is greater than 110 is obtained with a normal distribution of mean 100 and standard deviation 12 can be estimated using the z-table for Z = (110 -  100)/12 = 0.83So P (X > 110) = P (Z > 0.83) = 0.2033(b) Probability that a sample of 25 scores will have a mean greater than 105:we use a standard distribution with the same mean (100) but the standard distribution reduced by a factor of [tex]\sqrt{25} = 5[/tex]. That is a standard deviation of 12/5 = 2.4. which gives a Z-value of (105-100) / 2.4 = 2.08P (X> 105) = P (Z > 2.08) = 0.0188(c) Probability that a sample of 64 scores will have a mean greater than 105:we use a standard distribution with the same mean (100) but the standard deviation reduced by a factor of [tex]\sqrt{64} = 8[/tex]. That is a standard deviation of 12/8 = 1.5. which gives a Z-value of (105-100) / 1.5 = 3.33P (X> 105) = P (Z > 3.33) = 0.0004(d) the probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105This will be the addition of the two probabilities. We use a standard distribution with the same mean (100) but the standard distribution reduced by a factor of [tex]\sqrt{16} = 4[/tex]. That is a standard deviation of 12/4 = 3. which gives us two different Z values to study: (105-100) / 3 = 1.67and for X= 95 ==> Z = (95 - 100)/3 = - 1.67P (X > 105) = P (Z > 1.67) = 0.0475P (X < 95) = P (Z < -1.67) = 0.0475which add up to: 0.095.

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