random sample of n = 16 scores is selected from a normal distribution with a mean of μ = 50 and standard deviation of σ = 10.
a. what is the probability that the sample mean will have a value between 45 and 55?
b. what is the probability that the sample mean will have a value between 48 and 52?
c. what range of values contains the middle 95% of all possible sample means for n = 16?
Find z-scores using sample size:
z = (x - mean)/(sd/√n)
For a):
x = 45, 55
mean = 50
sd = 10
n = 16
Find two z-scores, using the values above. Use a z-table to find the probability between the two scores.
For b):
x = 48, 52
mean = 50
sd = 10
n = 16
Find two z-scores, using the values above. Use a z-table to find the probability between the two scores.
For c):
-1.96 = (x - 50)/(10/√16)
1.96 = (x - 50)/(10/√16)
Solve both equations for x. Those values will be your range of values containing the middle 95%.
I hope this will help get you started.
a. To find the probability that the sample mean will have a value between 45 and 55, we can use the Central Limit Theorem. According to the theorem, for a random sample from any population with mean μ and standard deviation σ, the sampling distribution of the sample mean approaches a normal distribution with mean μ and standard deviation σ/√n, where n is the sample size.
In this case, the population mean is μ = 50, the population standard deviation is σ = 10, and the sample size is n = 16.
First, we need to calculate the standard deviation of the sample mean, which is σ/√n = 10/√16 = 10/4 = 2.5.
Next, we standardize the values 45 and 55 to the z-scores by subtracting the mean and dividing by the standard deviation:
For 45: (45 - 50) / 2.5 = -2
For 55: (55 - 50) / 2.5 = 2
Now, we need to find the probability that the sample mean falls between -2 and 2 in the standard normal distribution. Using a z-table or calculator, the probability is approximately 0.9545.
Therefore, the probability that the sample mean will have a value between 45 and 55 is approximately 0.9545.
b. Similarly, to find the probability that the sample mean will have a value between 48 and 52, we can use the same method. The mean and standard deviation remain the same, but the values change.
Standardizing 48 and 52:
For 48: (48 - 50) / 2.5 = -0.8
For 52: (52 - 50) / 2.5 = 0.8
Now, we need to find the probability that the sample mean falls between -0.8 and 0.8 in the standard normal distribution. Using a z-table or calculator, the probability is approximately 0.6331.
Therefore, the probability that the sample mean will have a value between 48 and 52 is approximately 0.6331.
c. The range of values that contains the middle 95% of all possible sample means for n = 16 can be determined using the concept of a confidence interval.
Since the population standard deviation is known, we can use the z-distribution to find the critical values. The critical values are the z-values that correspond to the lower and upper percentages (2.5% each) of the distribution, which leave a middle 95% interval.
For a 95% confidence interval, we subtract and add the critical z-value to the sample mean:
Lower bound = x̄ - (z * σ/√n)
Upper bound = x̄ + (z * σ/√n)
Using a z-table or calculator, the critical z-value for a 95% confidence level is approximately 1.96.
Substituting the values:
Lower bound = 50 - (1.96 * 10/√16) ≈ 45.1
Upper bound = 50 + (1.96 * 10/√16) ≈ 54.9
Therefore, the range of values that contains the middle 95% of all possible sample means for n = 16 is approximately 45.1 to 54.9.
To answer these questions, we can use the Central Limit Theorem, which states that the sampling distribution of the sample mean approaches a normal distribution when the sample size is large enough.
a. To find the probability that the sample mean will have a value between 45 and 55, we can use the standard normal distribution. First, we need to calculate the standard error of the sample mean (also known as the standard deviation of the sampling distribution), which is equal to the population standard deviation divided by the square root of the sample size:
Standard Error = σ / √n
Standard Error = 10 / √16
Standard Error = 10 / 4
Standard Error = 2.5
Next, we need to convert the values of 45 and 55 to z-scores using the formula:
z = (x - μ) / Standard Error
For 45: z = (45 - 50) / 2.5 = -2
For 55: z = (55 - 50) / 2.5 = 2
Now we can use a standard normal distribution table or a calculator to find the probability associated with these z-scores. The probability that the sample mean will have a value between 45 and 55 is the difference between the cumulative probability up to 55 and the cumulative probability up to 45:
P(45 ≤ x ≤ 55) = P(z ≤ 2) - P(z ≤ -2)
Using the table or calculator, you can find these probabilities and subtract them to obtain the final result.
b. Similarly, to find the probability that the sample mean will have a value between 48 and 52, you need to calculate the z-scores for these values and use the same formula as in part a.
c. To determine the range of values that contains the middle 95% of all possible sample means for n = 16, we need to find the z-scores that correspond to the upper and lower tails of 2.5% (since 100% - 95% = 5% and we split it into two tails of 2.5% each).
To find the z-score associated with a tail of 2.5%, you can use a standard normal distribution table or calculator to find the z-score corresponding to a cumulative probability of 0.025 (lower tail) and 0.975 (upper tail). These z-scores represent the number of standard errors away from the mean.
Once you have the z-scores, you can use the formula:
Range of values = μ ± (z * Standard Error)
where μ is the population mean, z is the z-score, and Standard Error is the standard error of the sample mean. This will give you the lower and upper bounds of the range that encompasses the middle 95% of all possible sample means.