Herky found that 8 out of every 100 beans in his garden were nibbled by rabbits. how many beans can he expect to be damaged if t
2 answers:
First, find the unit rate, which is, out of how many beans will 1 be nibbled by a rabbit. We can do this by dividing the amount of beans by the amount that had been nibbled:
So one out of every 12.5 beans are nibbled on. There are 450 beans in his garden, we can divide our unit rate by this to find out how many he can expect to be damaged:
So he can expect 36 beans in his garden to be damaged.
So one out of every 12.5 beans are nibbled on. There are 450 beans in his garden, we can divide our unit rate by this to find out how many he can expect to be damaged:
So he can expect 36 beans in his garden to be damaged.
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Answer:
0.3594 = 35.94% probability that greater than 96 out of 153 DVDs will work correctly.
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:
So
Probability that greater than 96 out of 153 DVDs will work correctly.
97 or more DVDs, which is 1 subtracted by the pvalue of Z when X = 97. So
has a pvalue of 0.6406
1 - 0.6406 = 0.3594
0.3594 = 35.94% probability that greater than 96 out of 153 DVDs will work correctly.