4.Suppose you administered an anxiety test to a large sample of people and obtained normally distributed scores with a mean of 45 and standard deviation of 4. Do not use web-calculator to answer the following questions. Instead, you need to use the Z distribution table in Appendix A in Jackson’s book.
d.There are 200 students in a sample. How many of these students will have scores that fall under the score of 41?
The z-score for your 41 is
(41-45)/4 = -1
so use your chart to find the decimal for a standard deviation of -1 (should be .159 depending on the accuracy of your chart)
multiply that by 200
To answer this question, we need to use the Z-score formula:
Z = (X - μ) / σ
where:
Z is the Z-score,
X is the raw score,
μ is the mean, and
σ is the standard deviation.
In this case, the mean (μ) is 45 and the standard deviation (σ) is 4.
Now, let's calculate the Z-score for a score of 41:
Z = (41 - 45) / 4
Z = -1
We can now use the Z distribution table to find the proportion of scores below the Z-score of -1.
Looking up the Z-score of -1 in the table, we find that the proportion is 0.1587.
To find the number of students out of 200 who have scores below 41, we multiply the proportion by the total number of students:
Number of students = Proportion * Total number of students
Number of students = 0.1587 * 200
Number of students ≈ 31.74
Rounding to the nearest whole number, approximately 32 students will have scores below 41.
To find out how many students will have scores that fall under the score of 41, we need to determine the area under the normal distribution curve to the left of 41.
Here's how we can do that using the Z distribution table:
1. Calculate the z-score for a score of 41 using the formula:
z = (x - μ) / σ
where x is the score, μ is the mean, and σ is the standard deviation.
In this case, x = 41, μ = 45, and σ = 4.
z = (41 - 45) / 4
= -4/4
= -1
2. Look up the value of the z-score (-1) in the Z distribution table.
The table will give us the area under the curve to the left of the z-score.
3. Find the corresponding area in the table.
The Z distribution table provides areas for positive z-scores, so we need to look up the absolute value of -1.
In the table, the closest value to 1.00 is 0.8413.
4. Subtract the area from 0.5 to get the area to the left of -1.
Since the normal distribution is symmetric, the area to the left of -1 is the same as the area to the right of 1.
So, the area to the left of -1 is 0.5 - 0.8413 = 0.1587.
5. Multiply the area by the number of students in the sample (200) to find the number of students with scores below 41.
Number of students = area * sample size
= 0.1587 * 200
≈ 31.74
Therefore, approximately 31.74 students in the sample will have scores that fall under the score of 41.
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