Suppose that a random sample of size 36 is to be selected from a population with mean 50 and standard deviation 7. What is the approximate probability that will be within 0.5 of the population mean?
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Answer:
The probability that the sample mean will be within 0.5 of the population mean is 0.3328.
Step-by-step explanation:
It is provided that a random variable X has mean, μ = 50 andstandard deviation, σ = 7.
A random sample of size, n = 36 is selected.
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.
Then, the mean of the distribution of sample mean is given by,
And the standard deviation of the distribution of sample mean is given by,
So, the distribution of the sample mean of X is N (50, 1.167²).
Compute the probability that the sample mean will be within 0.5 of the population mean as follows:
Thus, the probability that the sample mean will be within 0.5 of the population mean is 0.3328.
Final answer:
To approximate the probability that the sample mean will be within 0.5 of the population mean, we can use the Central Limit Theorem. This theorem states that the sampling distribution of the sample means will be approximately normal, regardless of the shape of the population distribution, as long as the sample size is large enough. To calculate the probability, we need to find the standard error of the mean (SE), calculate the z-score for the upper bound of 0.5 deviations above the mean, and then find the cumulative probability corresponding to that z-score using a z-table or calculator.
Explanation:
To find the approximate probability that the sample mean will be within 0.5 of the population mean, we can use the Central Limit Theorem. According to the Central Limit Theorem, the sampling distribution of the sample means will be approximately normal, regardless of the shape of the population distribution, as long as the sample size is large enough (typically n ≥ 30).
- Calculate the standard error of the mean (SE) using the formula SE = σ/√n, where σ is the standard deviation of the population and n is the sample size. In this case, σ = 7 and n = 36, so SE = 7/√36 = 7/6 = 1.1667.
- Next, calculate the z-score corresponding to the upper bound of 0.5 deviations from the mean by using the formula z = (X - μ)/SE, where X is the value 0.5 deviations above the mean (50 + 0.5 = 50.5 in this case), μ is the mean of the population, and SE is the standard error of the mean. The z-score for 0.5 deviations above the mean can be calculated as z = (50.5 - 50)/1.1667 ≈ 0.4292.
- Finally, use a z-table or a calculator to find the probability corresponding to the z-score found in the previous step. The probability can be determined by subtracting the cumulative probability of the lower bound (z = -0.4292) from the cumulative probability of the upper bound (z = 0.4292). This can be expressed as p = P(Z < 0.4292) - P(Z < -0.4292).
Using a standard normal distribution table or a calculator, the approximate probability that the sample mean will be within 0.5 of the population mean is the difference between the cumulative probabilities of the upper and lower bounds found in step 3.
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Answer:
B and A
Step-by-step explanation:
(A question needs two answers so the asker can have the brainliest option)
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nahnahnahnahnahnahnah
What does the confidence interval suggest about the population means?
A. The confidence interval includes 0 which suggests that the two population means might be equal. There doesn't appear to be a significant difference between the mean drying time for paint type A and the mean drying time for paint type B. The modification does not seem to be effective in reducing drying times.
B. The confidence interval includes only positive values which suggests that the mean drying time for paint type A is greater than the mean drying time for paint type B. The modification seems to be effective in reducing drying times.
C. The confidence interval includes only positive values which suggests that the two population means might be equal. There doesn't appear to be a significant difference between the mean drying time for paint type A and the mean drying time for paint type B. The modification does not seem to be effective in reducing drying times.
D. The confidence interval includes only positive values which suggests that the mean drying time for paint type A is smaller than the mean drying time for paint type B. The modification does not seem to be effective in reducing drying times.
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This question is not complete, I got the complete one from google as below:
A paint manufacturer made a modification to a paint to speed up its drying time. Independent simple random samples of 11 cans of type A (the original paint) and 9 cans of type B (the modified paint) were selected and applied to similar surfaces. The drying times, in hours, were recorded.
The summary statistics are as follows.
Type A Type B
x1 = 76.3 hrs x2 = 65.1 hrs
s1 = 4.5 hrs s2 = 5.1 hrs
n1 = 11 n2 = 9
The following 98% confidence interval was obtained for μ1 - μ2, the difference between the mean drying time for paint cans of type A and the mean drying time for paint cans of type B:
4.90 hrs < μ1 - μ2 < 17.50 hrs
What does the confidence interval suggest about the population means?
A. The confidence interval includes 0 which suggests that the two population means might be equal. There doesn't appear to be a significant difference between the mean drying time for paint type A and the mean drying time for paint type B. The modification does not seem to be effective in reducing drying times.
B. The confidence interval includes only positive values which suggests that the mean drying time for paint type A is greater than the mean drying time for paint type B. The modification seems to be effective in reducing drying times.
C. The confidence interval includes only positive values which suggests that the two population means might be equal. There doesn't appear to be a significant difference between the mean drying time for paint type A and the mean drying time for paint type B. The modification does not seem to be effective in reducing drying times.
D. The confidence interval includes only positive values which suggests that the mean drying time for paint type A is smaller than the mean drying time for paint type B. The modification does not seem to be effective in reducing drying times.
Answer:
Option B is correct - the confidence interval includes only positive values which suggests that the mean drying time for paint type A is greater than the mean drying time for paint type B. The modification seems to be effective in reducing drying times.
Step-by-step explanation:
The 98% confidence interval for the difference in mean drying times of the two types of paints is (4.90, 17.50). This implies that Type A takes between 4.90 and 17.50 hours more to dry than type B paint.
Thus, option B is correct - the confidence interval includes only positive values which suggests that the mean drying time for paint type A is greater than the mean drying time for paint type B. The modification seems to be effective in reducing drying times.
Write an equation for the line in point-slope form.
(Simplify your answer. Use integers or fractions for any numbers in the equation.)
I WILL GIVE BRAINLIEST!!
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Answer:
Step-by-step explanation:
perp. -5
y + 7 = -5(x - 4)
y + 7 = -5x + 20
y = -5x + 13
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If you miss 8 out of 67 questions on the exam, your grade percentage would be approximately 88.06%.
Given that there are 67 questions on an exam. If you miss 8 of these questions then Subtract the number of questions missed (8) from the total number of questions (67):
Number of questions answered correctly = Total questions - Questions missed
= 67 - 8
= 59
Now, Grade percentage = (Number of questions answered correctly / Total questions) * 100
= (59 / 67) * 100
≈ 88.06% (rounded to two decimal places)
Therefore, if you miss 8 out of 67 questions on the exam, your grade percentage would be approximately 88.06%.
Learn more about percent here:
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Step-by-step explanation: