The mean of a set of normally distributed data is 600 with a standard devastation of 20. What percent of data is between 580 and
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Answer:
47.54% probability that more than 20 households but fewer than 35 households receive a retirement income
Step-by-step explanation:
We use the binomial aproxiation 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 a random sample of 120 households, what is the probability that more than 20 households but fewer than 35 households receive a retirement income?
We are working with discrete values, so this is the pvalue of Z when X = 35-1 = 34 subtracted by the pvalue of Z when X = 20 + 1 = 21.
X = 34
has a pvalue of 0.9993
X = 21
has a pvalue of 0.5239
0.9993 - 0.5239 = 0.4754
47.54% probability that more than 20 households but fewer than 35 households receive a retirement income