The number 21 is a(n) — number because
1 answer:
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Answer:
a) $5,656.85
b) Bell-shaped(normally distributed).
c) 36.32% probability of selecting a sample with a mean of at least $112,000.
d) 96.16% probability of selecting a sample with a mean of more than $100,000.
e) 59.84% probability of selecting a sample with a mean of more than $100,000 but less than $112,000.
Step-by-step explanation:
To solve this question, it is important to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are 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 random variable X, with mean and standard deviation
, a large sample size, of size at least 30, can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
a. If we select a random sample of 50 households, what is the standard error of the mean?
This is the standard deviation of the sample, that is, s, when .
So
b. What is the expected shape of the distribution of the sample mean?
By the Central Limit Theorem, bell-shaped(normally distributed).
c. What is the likelihood of selecting a sample with a mean of at least $112,000?
This is 1 subtracted by the pvalue of Z when X = 112000. So
By the Central Limit Theorem
has a pvalue of 0.6368
So 1-0.6368 = 0.3632 = 36.32% probability of selecting a sample with a mean of at least $112,000.
d. What is the likelihood of selecting a sample with a mean of more than $100,000?
This is 1 subtracted by the pvalue of Z when X = 112000. So
has a pvalue of 0.0384.
So 1-0.0384 = 0.9616 = 96.16% probability of selecting a sample with a mean of more than $100,000.
e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000
This is the pvalue of Z when X = 112000 subtractex by the pvalue of Z when X = 100000.
So
X = 112000
has a pvalue of 0.6368
X = 100000
has a pvalue of 0.0384.
So 0.6368 - 0.0384 = 0.5984 = 59.84% probability of selecting a sample with a mean of more than $100,000 but less than $112,000.