Assume Grades In A Course Follow A Normal Distribution

Assume Grades In A Course Follow A Normal Distribution The Average Gr

Assume grades in a course follow a normal distribution. The average grade in a class is 75, and the standard deviation is 5 points. A professor takes a sample of 10 people in the class and calculates the average for that class is also 75. In other words, the average grade of the sample is the experiment. What is the probability of getting an average of less than 75 from any sample?

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The problem involves understanding the properties of the sampling distribution of the sample mean when the population follows a normal distribution. Given that the population mean (μ) is 75 and the standard deviation (σ) is 5, the scenario considers selecting a sample of size n = 10 and calculating the sample mean (x̄). The question asks for the probability that the sample mean is less than 75, which, in this case, is the probability that the sample mean equals the population mean.

Since the population distribution is normal, and the sample size is small but greater than 1, the sampling distribution of the mean is also normal, with a mean equal to the population mean and a standard error (SE) computed as σ/√n. The standard error (SE) is calculated as:

SE = σ / √n = 5 / √10 ≈ 5 / 3.162 ≈ 1.58

The Z-score corresponding to the sample mean of 75 is calculated by:

Z = (x̄ - μ) / SE = (75 - 75) / 1.58 = 0 / 1.58 = 0

Because the Z-score is zero, the probability of observing a sample mean less than 75 is equivalent to the probability that Z

This result aligns with the symmetry property of the normal distribution. The mean of the sampling distribution coincides with the population mean, and the probability of the sample mean falling below or above this value is equally likely, assuming random sampling.

In conclusion, the probability of obtaining a sample mean less than 75 when sampling from a normally distributed population with mean 75 and standard deviation 5, with a sample size of 10, is approximately 50%. This outcome demonstrates the fundamental properties of normal distributions and sampling variability in statistical inference.

Part IV: Question 1a & 1b (Non-SPSS)

Age at onset of dementia was determined for a sample of adults between the ages of 60 and 75. For a sample of 20 subjects, M = 69.8 and s = 2.79. Use this information to answer the following:

1-a) Based on the data you have and the z table, what percentage of people might start to show signs of dementia at or before age 65?

First, calculate the Z-score for age 65:

Z = (X - M) / s = (65 - 69.8) / 2.79 ≈ -4.8 / 2.79 ≈ -1.72

Referring to the standard normal distribution table, a Z-score of -1.72 corresponds to a cumulative probability of approximately 0.0427, or 4.27%. This indicates that about 4.27% of the population might begin showing signs of dementia at or before age 65.

1-b) If you consider the normal range of onset in this population to be +/-1 z-score from the mean, what two ages correspond to this? Answer (+1 z)

Calculate the upper and lower bounds:

Lower age (corresponding to -1 z-score):

Age = M + (Z s) = 69.8 + (-1 2.79) = 69.8 - 2.79 = 67.01

Upper age (corresponding to +1 z-score):

Age = 69.8 + (1 * 2.79) = 69.8 + 2.79 = 72.59

Thus, the normal range of dementia onset ages is approximately from 67.01 to 72.59 years.

Part IV: Question 2a-2b (Non-SPSS)

For each of the following scenarios, compute the effect size and state whether it is approximately small, medium, or large.

2-a) Participants in an inpatient study of treatment for OCD complete an anxious feelings inventory which is then compared to the general population. The sample mean is M = 26.2. The mean in the general population of inpatients on this inventory is 35.1, and the population standard deviation is 1.5.

Calculate the effect size (Cohen's d):

d = (M_sample - M_population) / σ_population = (26.2 - 35.1) / 1.5 ≈ -8.9 / 1.5 ≈ -5.93

The absolute value of the effect size is approximately 5.93, which is considered a very large effect according to Cohen's conventions (

2-b) A mood assessment in a sample of 15 gym members has a mean of 83 and a standard deviation of 5.6. The mean in the general population on this measure is 79.5.

Calculate the effect size (Cohen's d):

d = (83 - 79.5) / 5.6 ≈ 3.5 / 5.6 ≈ 0.625

This effect size approximates a medium effect size based on Cohen's criteria (

References

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