Out of 2000 random but normally distributed numbers with a mean of 45 and a standard deviation of x, approximately 1360 numbers
1 answer:
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Answer:
71.08% probability that pˆ takes a value between 0.17 and 0.23.
Step-by-step explanation:
We use the binomial approxiation to the normal to solve this question.
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:
. So
In other words, find probability that pˆ takes a value between 0.17 and 0.23.
This probability is the pvalue of Z when X = 200*0.23 = 46 subtracted by the pvalue of Z when X = 200*0.17 = 34. So
X = 46
has a pvalue of 0.8554
X = 34
has a pvalue of 0.1446
0.8554 - 0.1446 = 0.7108
71.08% probability that pˆ takes a value between 0.17 and 0.23.