Answer :
Answer:
(a) 0.06681
(b) 21.600
Step-by-step explanation:
We have been given that the distribution of scores on a recent test closely followed a Normal Distribution with a mean of 22 points and a standard deviation of 2 points.
(a) We will use z-score formula to find the proportion of the students, who scored at least 25 points on this test.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
[tex]z[/tex] = Z-score,
[tex]x[/tex] = Sample score,
[tex]\mu[/tex] = Mean,
[tex]\sigma[/tex] = Standard deviation.
[tex]z=\frac{25-22}{2}[/tex]
[tex]z=\frac{3}{2}[/tex]
[tex]z=1.5[/tex]
Now, we will need to find the area under normal distribution curve such that [tex]p(z\geq1.5)[/tex].
[tex]p(z\geq a)=1-p(z< a)[/tex]
Using normal distribution table, we will get:
[tex]p(z\geq1.5)=1-p(z<1.5)[/tex]
[tex]p(z\geq1.5)=1-0.93319 [/tex]
[tex]p(z\geq1.5)=0.06681[/tex]
Therefore, the proportion of the students, who scored at least 25 points on this test is 0.06681.
(b) To find 42 percentile of the distribution of test scores, we will find z-score corresponding to 42% using normal distribution table.
From normal distribution table we get z-score is [tex]-0.20[/tex].
We will substitute this value is z-score formula and solve for x.
[tex]z=\frac{x-\mu}{\sigma}[/tex]
[tex]-0.20=\frac{x-22}{2}[/tex]
[tex]-0.20*2=\frac{x-22}{2}*2[/tex]
[tex]-0.40=x-22[/tex]
[tex]-0.40+22=x-22+22[/tex]
[tex]21.6=x[/tex]
Therefore, 21.600 is the 42 percentile of the distribution of test scores.
6.68% of students scored at least 25 points on this test
Z score
The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
z = (x - μ)/σ
where x is raw score, σ is standard deviation and μ is mean
μ = 22, σ = 2
For x > 25:
z = (25 - 22)/2 = 1.5
P(z > 1.5) = 1 - P(z < 1.5) = 1 - 0.9332 = 0.0668
42 percentile has a z score of -0.2:
-0.2 = (x - 22)/2
x = 21.6
6.68% of students scored at least 25 points on this test
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