Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution. In one county, the conviction rate for speeding is 85%. Estimate the probability that of the next 100 speeding summonses issued, there will be at least 90 convictions.

13.14% probability that of the next 100 speeding summonses issued, there will be at least 90 convictions.

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

Estimate the probability that of the next 100 speeding summonses issued, there will be at least 90 convictions.

At least 90 is more than 89, which is 1 subtracted by the pvalue of Z when X = 89. So

has a pvalue of 0.8686

1 - 0.8686 = 0.1314

13.14% probability that of the next 100 speeding summonses issued, there will be at least 90 convictions.

300pi

step-by-step explanation: