the population of sat scores forms a normal distribution with a mean of u=500 and a standard deviation of o=100. if the average sat score is 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>510)?
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 between m=510 and m=520? in symbols, what is p(510<m<520)?
Z = (score-mean)/SEm
SEm = SD/√n
Treat the new mean as a score.
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 need to use the concept of the sampling distribution of the sample mean. The sampling distribution refers to the distribution of all possible sample means that can be obtained from a population.
In this case, we know that the population of SAT scores follows a normal distribution with a mean (μ) of 500 and a standard deviation (σ) of 100.
When sampling from a normal distribution, the distribution of the sample means will also be normal, and its mean (μ) will be the same as the population mean. However, the standard deviation of the sample means, also known as the standard error (σ/√n), will be smaller than the population standard deviation, where n is the sample size.
Now, let's answer the questions step by step:
a. To find the probability that the sample mean will be greater than 510 (p(m>510)), we need to calculate the z-score for 510 using the formula: z = (X - μ) / (σ/√n). Plugging in the values, we have z = (510 - 500) / (100 / √25) = 10 / 20 = 0.5.
Next, we can refer to a standard normal distribution table or use statistical software to find the probability associated with the z-score of 0.5. Looking up the probability of a z-score being greater than 0.5, we find it to be approximately 0.3085.
Therefore, the probability (p) that the sample mean will be greater than m=510 is p(m>510) = 0.3085.
b. Using the same steps as in part a, we find the z-score for 520 to be z = (520 - 500) / (100 / √25) = 20 / 20 = 1.
Looking up the probability of a z-score being greater than 1, we find it to be approximately 0.1587.
Therefore, the probability (p) that the sample mean will be greater than m=520 is p(m>520) = 0.1587.
c. To find the probability that the sample mean will be between 510 and 520 (p(510<m<520)), we need to find the probability that the sample mean is less than 520 minus the probability that the sample mean is less than 510.
Using the z-score approach, we find the z-score for 520 to be 1 (as calculated in part b), and the z-score for 510 to be (510 - 500) / (100 / √25) = 10 / 20 = 0.5 (calculated in part a).
Now, we can find the probabilities associated with these z-scores:
For z = 1, the probability is approximately 0.8413.
For z = 0.5, the probability is approximately 0.6915.
To find p(510<m<520), we subtract the probability associated with 510 from the probability associated with 520: p(510<m<520) = 0.8413 - 0.6915 = 0.1498.
Therefore, the probability (p) that the sample mean will be between m=510 and m=520 is p(510<m<520) = 0.1498.
To solve these problems, we will use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means will be approximately normally distributed, regardless of the shape of the population distribution, as long as the sample size is large enough (usually considered to be greater than 30).
a. To find the probability that the sample mean will be greater than 510 (i.e., P(m > 510)), we need to calculate the z-score and then look up the corresponding probability from the standard normal distribution.
First, we calculate the z-score:
z = (510 - μ) / (σ / √n)
= (510 - 500) / (100 / √25)
= 10 / (100 / 5)
= 10 / 20
= 0.5
Next, we can use a z-table or a calculator to find the probability associated with a z-score of 0.5. Looking up the z-score in the table, we find that the probability is approximately 0.6915.
Therefore, P(m > 510) ≈ 0.6915.
b. Similarly, to find the probability that the sample mean will be greater than 520 (i.e., P(m > 520)), we calculate the z-score:
z = (520 - μ) / (σ / √n)
= (520 - 500) / (100 / √25)
= 20 / (100 / 5)
= 20 / 20
= 1
Using the z-table, we find that the probability associated with a z-score of 1 is approximately 0.8413.
Therefore, P(m > 520) ≈ 0.8413.
c. To find the probability that the sample mean will be between 510 and 520 (i.e., P(510 < m < 520)), we need to calculate the z-scores for the lower and upper bounds.
For the lower bound:
z_lower = (510 - μ) / (σ / √n)
= (510 - 500) / (100 / √25)
= 10 / (100 / 5)
= 0.5
For the upper bound:
z_upper = (520 - μ) / (σ / √n)
= (520 - 500) / (100 / √25)
= 20 / (100 / 5)
= 1
Using the z-table, we find that the probability associated with a z-score of 0.5 is approximately 0.6915, and the probability associated with a z-score of 1 is approximately 0.8413.
Therefore, P(510 < m < 520) ≈ P(m > 510) - P(m > 520)
≈ 0.6915 - 0.8413
≈ -0.1498
However, a negative probability does not make sense in this context. Instead, the correct value should be the absolute value of the difference, which gives us approximately 0.1498.
Therefore, P(510 < m < 520) ≈ 0.1498.