The population of SAT scores forms a normal distribution with a mean of µ =500 and a standard deviation of σ =100. If the average SAT score calculated for a sample of n = 25 students,
a. What is the probability that the sample mean will be greater than M= 510. In symbols, what is p (M >520?
b. What is the probability that the sample mean will be greater than M = 520. In symbols, what is p (M>520?
c. What is the probability that the sample mean will be between M =510 and M = 520? In symbols, what is p (510 < M < 520?
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.
To answer these questions, we will use the Central Limit Theorem and assume that the population of SAT scores is normally distributed.
a. What is the probability that the sample mean will be greater than M = 510? In symbols, what is p (M > 520)?
To calculate this probability, we need to find the z-score for the sample mean M = 510. The formula for the z-score is:
z = (x - µ) / (σ / √n)
where x is the sample mean, µ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, x = 510, µ = 500, σ = 100, and n = 25. Plugging these values into the formula, we get:
z = (510 - 500) / (100 / √25)
= 10 / (100 / 5)
= 10 / 20
= 0.5
Now, we need to find the probability corresponding to this z-score using a standard normal distribution table or a calculator. The probability of the sample mean being greater than 520 can be calculated as:
p (M > 520) = 1 - p (M ≤ 520)
= 1 - P(z ≤ 0.5)
Referring to a standard normal distribution table or using a calculator, we find that P(z ≤ 0.5) is approximately 0.6915. So:
p (M > 520) = 1 - 0.6915
= 0.3085
Therefore, the probability that the sample mean will be greater than 520 is approximately 0.3085.
b. What is the probability that the sample mean will be greater than M = 520? In symbols, what is p (M > 520)?
Using the same approach as in part (a), we can find the z-score for M = 520:
z = (520 - 500) / (100 / √25)
= 20 / (100 / 5)
= 20 / 20
= 1
Now we find the probability corresponding to this z-score using a standard normal distribution table or a calculator:
p (M > 520) = 1 - P(z ≤ 1)
Looking up in the standard normal distribution table or using a calculator, we find that P(z ≤ 1) is approximately 0.8413. So:
p (M > 520) = 1 - 0.8413
= 0.1587
Therefore, the probability that the sample mean will be greater than 520 is approximately 0.1587.
c. What is the probability that the sample mean will be between M = 510 and M = 520? In symbols, what is p (510 < M < 520)?
To find this probability, we need to calculate the z-scores for M = 510 and M = 520, and then find the probability between these two z-scores.
The z-score for M = 510:
z1 = (510 - 500) / (100 / √25)
= 10 / (100 / 5)
= 10 / 20
= 0.5
The z-score for M = 520:
z2 = (520 - 500) / (100 / √25)
= 20 / (100 / 5)
= 20 / 20
= 1
Now, we find the probability between these two z-scores using a standard normal distribution table or a calculator:
p (510 < M < 520) = P(z1 < z < z2)
= P(0.5 < z < 1)
Referring to a standard normal distribution table or using a calculator, we find that P(0.5 < z < 1) is approximately 0.3413.
Therefore, the probability that the sample mean will be between 510 and 520 is approximately 0.3413.
To answer these questions, we need to use the concept of the sampling distribution of the sample mean. The sampling distribution of the sample mean refers to the distribution of sample means that would be obtained if we repeatedly sampled from a population and calculated the mean of each sample.
a. To find the probability that the sample mean will be greater than 520 (p(M > 520)), we need to calculate the z-score and then find the corresponding probability from the standard normal distribution.
The formula to calculate the z-score is:
z = (M - µ) / (σ / √n)
In this case, M = 520, µ = 500, σ = 100, and n = 25.
Plugging in the values, we have:
z = (520 - 500) / (100 / √25)
z = 20 / (100 / 5)
z = 20 / 20
z = 1
Now, we need to find the probability associated with a z-score of 1. Using a standard normal distribution table or a calculator, we find that the probability is 0.8413.
Therefore, p(M > 520) = 0.8413.
b. Similar to part a, to find the probability that the sample mean will be greater than 520 (p(M > 520)), we use the same formula:
z = (M - µ) / (σ / √n)
In this case, M = 520, µ = 500, σ = 100, and n = 25.
Plugging in the values, we have:
z = (520 - 500) / (100 / √25)
z = 20 / (100 / 5)
z = 20 / 20
z = 1
Since we already calculated the probability associated with a z-score of 1 in part a, we can conclude that p(M > 520) = 0.8413.
c. To find the probability that the sample mean will be between 510 and 520 (p(510 < M < 520)), we need to find the difference between the probabilities of two z-scores.
We calculate the z-scores for 510 and 520 using the same formula:
For 510:
z1 = (510 - 500) / (100 / √25)
z1 = 10 / (100 / 5)
z1 = 10 / 20
z1 = 0.5
For 520:
z2 = (520 - 500) / (100 / √25)
z2 = 20 / (100 / 5)
z2 = 20 / 20
z2 = 1
Using a standard normal distribution table or a calculator, we find that the probability associated with z1 = 0.5 is 0.6915 and the probability associated with z2 = 1 is 0.8413.
Therefore, p(510 < M < 520) = 0.8413 - 0.6915 = 0.1498.