one fourth of the 7th grade students made honor roll in the first quarter. the number of honor roll students increased by 18 in
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Answer:
a) 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.
b) 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.
Step-by-step explanation:
To solve this question, it is important to know the Normal probability distribution and the Central Limit Theorem
Normal probability distribution
Problems of normally distributed samples 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 random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
In this problem, we have that:
A) What is the chance HLI will find a sample mean between 5.5 and 7.1 hours?
This is the pvalue of Z when X = 7.1 subtracted by the pvalue of Z when X = 5.5.
By the Central Limit Theorem, the formula for Z is:
X = 7.1
has a pvalue of 0.9962
X = 5.5
has a pvalue of 0.0038
So there is a 0.9962 - 0.0038 = 0.9924 = 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.
B) Calculate the probability that the sample mean will be between 5.9 and 6.7 hours.
This is the pvalue of Z when X = 6.7 subtracted by the pvalue of Z when X = 5.9
X = 6.7
has a pvalue of 0.9082
X = 5.9
has a pvalue of 0.0918.
So there is a 0.9082 - 0.0918 = 0.8164 = 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.