A state insurance commission estimates that 13% of all motorists in its state areuninsured. Suppose this proportion is valid. Find the probability that in a randomsample of 50 motorists, at least 5 will be uninsured. You may assume that the normaldistribution applies.

79.95% probability that in a randomsample of 50 motorists, at least 5 will be uninsured.

Step-by-step explanation:

We use the binomial approximation 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

Find the probability that in a randomsample of 50 motorists, at least 5 will be uninsured.

This is 1 subtracted by the pvalue of Z when X = 4. So

has a pvalue of 0.2005

1 - 0.2005 = 0.7995

79.95% probability that in a randomsample of 50 motorists, at least 5 will be uninsured.

2/5

step-by-step explanation:

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐