How do you divide 360 into 6
1 answer:
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Answer:
a) s = 0.4
b) Z = 2.5
c) 99th percentile.
d) The larger sample size would lead to a smaller margin of error, which would lead to a higher z-score and a increased percentile.
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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes 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.
Assume the average height for American women is 64 inches with a standard deviation of 2 inches
This means that
A) Calculate the standard error for the distribution of means.
Sample of 25 means that , so
.
B) Calculate the z statistic for the sorority group.
Sample mean of 65 means that .
By the Central Limit Theorem
C) What is the approximate percentile for this sample? Enter as a whole number.
Z = 2.5 has a p-value of 0.9938, so 0.99*100 = 99th percentile.
D) If the sorority actually had 36 members (still with an average of 65 inches), would you expect the percentile value to increase or decrease? why?
The larger sample size would lead to a smaller margin of error, which would lead to a higher z-score and a increased percentile.