Nuclear power plants have redundant components in important systems to reduce the chance of catastrophic failure. Suppose that a plant has three gauges to measure the level of the coolant in the reactor core and each gauge has a 0.02 probability of failing. What is the probability that none of the gauges fails

0.9412 = 94.12% probability that none of the gauges fails

Step-by-step explanation:

For each gauge, there are only two possible outcomes. Either it fails, or it does not. The probability of a gauge failing is independent of other gauges. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

3 gouges, each with a 0.02 probability of failing.

This means that

What is the probability that none of the gauges fails?

This is P(X = 0).

0.9412 = 94.12% probability that none of the gauges fails

*The dot plot is shown in the attachment below

2

Step-by-step explanation:

Interquartile range is the difference between the upper median (Q3) and the lower median (Q1).

First, let's write out each value given in the data. Each dot represents a data point.

We have:

2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 7

=>Find the median:

Our median is the middle value. The middle value is the 6th value = 4

==>Upper median Q3) = the middle value of the set of data we have from the median to our far right.

2, 3, 3, 4, 4, |4,| 4, 5, [5], 6, 7

Our upper median = 5

==>Lower median(Q1) = the middle value of the data set we have from our median to our far left.

2, 3, [3], 4, 4, |4,| 4, 5, 5, 6, 7

Lower median = 3

==>Interquartile range = Q3 - Q1 = 5-3 = 2