Respuesta :
Answer:
(a) 0.2033
(b) 0.0188
(c) 0.0004
(d) 0.095
Step-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 = 0.83
So 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 = 2.08
P (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 = 3.33
P (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 105
This 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 = 1.67
and for X= 95 ==> Z = (95 - 100)/3 = - 1.67
P (X > 105) = P (Z > 1.67) = 0.0475
P (X < 95) = P (Z < -1.67) = 0.0475
which add up to: 0.095.
Using the normal distribution and the central limit theorem, it is found that there is a:
a) 0.2033 = 20.33% probability that a single score drawn at random will be greater than 110.
b) 0.0188 = 1.88% probability that a sample of 25 scores will have a mean greater than 105.
c) 0.0004 = 0.04% probability that a sample of 64 scores will have a mean greater than 105.
d) 0.095 = 9.5% probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- Mean of 100, hence [tex]\mu = 100[/tex].
- Standard deviation of 12, hence [tex]\sigma = 12[/tex].
Item a:
This probability is 1 subtracted by the p-value of Z when X = 110, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{110 - 100}{12}[/tex]
[tex]Z = 0.83[/tex]
[tex]Z = 0.83[/tex] has a p-value of 0.7967.
1 - 0.7967 = 0.2033
0.2033 = 20.33% probability that a single score drawn at random will be greater than 110.
Item b:
Sample of 25, hence [tex]n = 25, s = \frac{12}{\sqrt{25}} = 2.4[/tex].
This probability is 1 subtracted by the p-value of Z when X = 105, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{105 - 100}{2.4}[/tex]
[tex]Z = 2.08[/tex]
[tex]Z = 2.08[/tex] has a p-value of 0.9812.
1 - 0.9812 = 0.0188
0.0188 = 1.88% probability that a sample of 25 scores will have a mean greater than 105.
Item c:
Sample of 64, hence [tex]n = 64, s = \frac{12}{\sqrt{64}} = 1.5[/tex].
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{105 - 100}{1.5}[/tex]
[tex]Z = 3.33[/tex]
[tex]Z = 3.33[/tex] has a p-value of 0.9996.
1 - 0.9996 = 0.0004
0.0004 = 0.04% probability that a sample of 64 scores will have a mean greater than 105.
Item d:
Sample of 16, hence [tex]n = 16, s = \frac{12}{\sqrt{16}} = 3[/tex].
Both 105 and 95 are the same distance of the mean, so we find one probability, and multiply by 2.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{105 - 100}{3}[/tex]
[tex]Z = 1.67[/tex]
[tex]Z = 1.67[/tex] has a p-value of 0.9525.
1 - 0.9525 = 0.0475.
2 x 0.0475 = 0.095
0.095 = 9.5% probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105.
A similar problem is given at https://brainly.com/question/24663213
