IQ scores form a normal distribution with a mean of μ = 100 and a standard deviation of σ = 15. What is the probability of obtaining a sample mean greater than M = 103? for a random sample of n = 9 people? for a random sample of n = 25 people? for a random sample of n = 100 people
The Correct Answer and Explanation is:
To find the probability of obtaining a sample mean greater than M=103M = 103M=103 for different sample sizes, we can use the Z-score formula for sample means: Z=M−μσnZ = \frac{M – \mu}{\frac{\sigma}{\sqrt{n}}}Z=nσM−μ
Where:
- MMM is the sample mean of interest (103),
- μ\muμ is the population mean (100),
- σ\sigmaσ is the population standard deviation (15),
- nnn is the sample size.
We will calculate the Z-score for each sample size and then use the standard normal distribution table to find the corresponding probability.
For n=9n = 9n=9:
First, calculate the standard error of the mean: SE=σn=159=153=5SE = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{9}} = \frac{15}{3} = 5SE=nσ=915=315=5
Now, calculate the Z-score: Z=103−1005=35=0.6Z = \frac{103 – 100}{5} = \frac{3}{5} = 0.6Z=5103−100=53=0.6
Using the Z-table, the probability corresponding to a Z-score of 0.6 is approximately 0.7257. This gives the probability of obtaining a sample mean less than 103. To find the probability of obtaining a sample mean greater than 103, subtract this value from 1: P(M>103)=1−0.7257=0.2743P(M > 103) = 1 – 0.7257 = 0.2743P(M>103)=1−0.7257=0.2743
For n=25n = 25n=25:
Calculate the standard error of the mean: SE=σn=1525=155=3SE = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{25}} = \frac{15}{5} = 3SE=nσ=2515=515=3
Now, calculate the Z-score: Z=103−1003=33=1Z = \frac{103 – 100}{3} = \frac{3}{3} = 1Z=3103−100=33=1
From the Z-table, the probability corresponding to a Z-score of 1 is approximately 0.8413. To find the probability of obtaining a sample mean greater than 103: P(M>103)=1−0.8413=0.1587P(M > 103) = 1 – 0.8413 = 0.1587P(M>103)=1−0.8413=0.1587
For n=100n = 100n=100:
Calculate the standard error of the mean: SE=σn=15100=1510=1.5SE = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{100}} = \frac{15}{10} = 1.5SE=nσ=10015=1015=1.5
Now, calculate the Z-score: Z=103−1001.5=31.5=2Z = \frac{103 – 100}{1.5} = \frac{3}{1.5} = 2Z=1.5103−100=1.53=2
From the Z-table, the probability corresponding to a Z-score of 2 is approximately 0.9772. To find the probability of obtaining a sample mean greater than 103: P(M>103)=1−0.9772=0.0228P(M > 103) = 1 – 0.9772 = 0.0228P(M>103)=1−0.9772=0.0228
Summary of Probabilities:
- For n=9n = 9n=9, the probability is 0.2743.
- For n=25n = 25n=25, the probability is 0.1587.
- For n=100n = 100n=100, the probability is 0.0228.
Explanation:
The probability of obtaining a sample mean greater than 103 decreases as the sample size increases because larger sample sizes result in a smaller standard error. As a result, the sample means cluster more closely around the population mean, making extreme values (like 103) less likely.
