IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability that a random sample of 20 people have a mean IQ score greater than 110? (Round to 4 decimal places)
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of that Z score.
To find the probability that a random sample of 20 people have a mean IQ score greater than 110, we can use the Central Limit Theorem, which states that for a large enough sample size, the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the population distribution.
First, we need to calculate the standard deviation of the sampling distribution, also known as the standard error of the mean (SEM). We can calculate the SEM as the population standard deviation divided by the square root of the sample size:
SEM = standard deviation / √sample size
SEM = 15 / √20 ≈ 3.3541
Next, we can calculate the z-score, which represents the number of standard deviations a particular value is from the mean. The z-score can be calculated using the formula:
z = (sample mean - population mean) / SEM
z = (110 - 100) / 3.3541 ≈ 2.9814
We can then use a standard normal distribution table or a calculator to find the probability associated with this z-score. The probability that a random sample of 20 people have a mean IQ score greater than 110 is the area under the standard normal distribution curve to the right of the z-score.
Using a standard normal distribution table or calculator, the probability corresponding to a z-score of 2.9814 is approximately 0.0014.
Therefore, the probability that a random sample of 20 people have a mean IQ score greater than 110 is approximately 0.0014, or 0.14% (rounded to 4 decimal places).
To find the probability that a random sample of 20 people have a mean IQ score greater than 110, we can use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means will be approximately normal, regardless of the shape of the original population, as long as the sample size is sufficiently large.
The mean of the sample means will also be equal to the population mean, which in this case is 100. The standard deviation of the sample means, known as the standard error, is equal to the population standard deviation divided by the square root of the sample size. In this case, the standard error is equal to 15 divided by the square root of 20.
First, we need to convert the value of 110 to a z-score. The z-score measures how many standard deviations an individual score is from the mean. We can calculate the z-score using the formula: z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation.
Applying the formula: z = (110 - 100) / (15 / √20) = 2.1082 (rounded to 4 decimal places).
Next, we can use a standard normal distribution table or a calculator with a standard normal distribution function to find the corresponding probability. Since we are interested in the probability that the mean IQ score is greater than 110, we need to find the area to the right of the z-score of 2.1082.
Using a standard normal distribution table or calculator, we find that the probability to the right of the z-score 2.1082 is 0.0178 (rounded to 4 decimal places).
Therefore, the probability that a random sample of 20 people have a mean IQ score greater than 110 is 0.0178 (rounded to 4 decimal places).