My sister needs help w this but i don’t understand lol
2 answers:
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Answer:
a) 99% of the sample means will fall between 0.933 and 0.941.
b) By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
(a) If the true mean is 0.9370 with a standard deviation of 0.0090 within what interval will 99% of the sample means fail?
Samples of 34 means that
We have that
By the Central Limit Theorem,
Within what interval will 99% of the sample means fail?
Between the (100-99)/2 = 0.5th percentile and the (100+99)/2 = 99.5th percentile.
0.5th percentile:
X when Z has a pvalue of 0.005. So X when Z = -2.575.
By the Central Limit Theorem
99.5th percentile:
X when Z has a pvalue of 0.995. So X when Z = 2.575.
99% of the sample means will fall between 0.933 and 0.941.
(b) If the true mean 0.9370 with a standard deviation of 0.0090, what is the sampling distribution of ¯X?
By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.