Can anyone solve this?
1 answer:
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Answer:
0.0037 = 0.37% probability that the home team would win 65% or more of its games in a simple random sample of 80 games
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.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean and standard deviation
The home team therefore wins 50% of its games
This means that
Determine the probability that the home team would win 65% or more of its games in a simple random sample of 80 games
Sample of 80 means that and, by the Central Limit Theorem:
This probability is 1 subtracted by the pvalue of Z when X = 0.65. So
By the Central Limit Theorem
has a pvalue of 0.9963
1 - 0.9963 = 0.0037
0.0037 = 0.37% probability that the home team would win 65% or more of its games in a simple random sample of 80 games