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Answer:
The mean of the sampling distribution of means for SRS of size 50 is and the standard deviation is
By the Central Limit Theorem, since we have of sample of 50, which is larger than 30, it does not matter that the underlying population distribution is not normal.
0% probability a sample of 50 people will wait longer than 45 seconds for an elevator.
Step-by-step explanation:
To solve this problem, we need 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 at least 30, can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
What are the mean and standard deviation of the sampling distribution of means for SRS of size 50?
By the Central Limit Theorem
The mean of the sampling distribution of means for SRS of size 50 is and the standard deviation is
Does it matter that the underlying population distribution is not normal?
By the Central Limit Theorem, since we have of sample of 50, which is larger than 30, it does not matter that the underlying population distribution is not normal.
What is the probability a sample of 50 people will wait longer than 45 seconds for an elevator?
We have to use 45 seconds as minutes, since the mean and the standard deviation are in minutes.
Each minute has 60 seconds.
So 45 seconds is 45/60 = 0.75 min.
This probability is 1 subtracted by the pvalue of Z when X = 0.75. So
By the Central Limit Theorem
has a pvalue of 1
1-1 = 0
0% probability a sample of 50 people will wait longer than 45 seconds for an elevator.
Answer:
a) 68.26% probability that a student scores between 350 and 550
b) A score of 638(or higher).
c) The 60th percentile of test scores is 475.3.
d) The middle 30% of the test scores is between 411.5 and 488.5.
Step-by-step explanation:
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.
In this problem, we have that:
a. What is the probability that a student scores between 350 and 550?
This is the pvalue of Z when X = 550 subtracted by the pvalue of Z when X = 350. So
X = 550
has a pvalue of 0.8413
X = 350
has a pvalue of 0.1587
0.8413 - 0.1587 = 0.6826
68.26% probability that a student scores between 350 and 550
b. If the upper 3% scholarship, what score must a student receive to get a scholarship?
100 - 3 = 97th percentile, which is X when Z has a pvalue of 0.97. So it is X when Z = 1.88
A score of 638(or higher).
c. Find the 60th percentile of the test scores.
X when Z has a pvalue of 0.60. So it is X when Z = 0.253
The 60th percentile of test scores is 475.3.
d. Find the middle 30% of the test scores.
50 - (30/2) = 35th percentile
50 + (30/2) = 65th percentile.
35th percentile:
X when Z has a pvalue of 0.35. So X when Z = -0.385.
65th percentile:
X when Z has a pvalue of 0.35. So X when Z = 0.385.
The middle 30% of the test scores is between 411.5 and 488.5.