A large electronic office product contains 2000 electronic components. Assume that the probability that each component operates without failure during the useful life of the product is 0.995, and assume that the components fail independently. Approximate the probability that 5 or more of the original 2000 components fail during the useful life of the product.
(apply the ±½ correction factor and round value of standard normal random variable to 2 decimal places.) Round your answer to four decimal places (e. g. 98.7654).

We can use the z score formula given by:

And using this formula we have:

And using the complement rule we have this:

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

The probability mass function for the Binomial distribution is given as:

Where (nCx) means combinatory and it's given by this formula:

And we want to find this probability:

For this case since we have independence and the following two conditions:

We can use the normal approximation from the binomial distribution, and the mean would be given by:

And using the continuity correction factor we have this:

We can use the z score formula given by:

And using this formula we have:

And using the complement rule we have this:

what are the options? ?

step-by-step explanation:

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.