The monthly income of Mr. Sharma is Rs. 16500.He

In a math class with 21 students, a test was given the same day that an assignment was due. There were 12 students who passed th

snow_lady [41]

Answer:

5/9

Step-by-step explanation:

The given is students that failed the test. 5 students completed the homework and failed the test and there is a total of 9 students who failed the test. So, the answer would be 5/9

4 0

3 years ago

Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 12

melamori03 [73]

Answer:

98.75% probability that every passenger who shows up can take the flight

Step-by-step explanation:

For each passenger who show up, there are only two possible outcomes. Either they can take the flight, or they do not. The probability of a passenger taking the flight is independent from other passenger. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.

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:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .