The first term of a sequence is 5.
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Answer:
120 has a z-score higher than 2.5. So yes, it would be unusual to have more than 120 successes out of 200 trials.
Step-by-step explanation:
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
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.
When we are approximating a binomial distribution to a normal one, we have that ,
.
In this problem, we have that:
Use the criterion that it is unusual to have data values more than 2.5 standard deviations above the mean or 2.5 standard deviations below the mean
This means that z-scores higher than 2.5 or lower than -2.5 are considered unusual.
Would it be unusual to have more than 120 successes out of 200 trials
We have to find the Z-score of X = 120.
So
120 has a z-score higher than 2.5. So yes, it would be unusual to have more than 120 successes out of 200 trials.