Mathematics, 17.06.2021 21:50 ToriChristine

The quality assurance engineer of a receiving-sets manufacturer inspects receiving-sets in lots of 50. He selects 4 of the 50 receiving-sets at random and inspects them thoroughly. Assuming that 5 of the 50 receiving-sets in the current lot are defective, find the probability that exactly 2 of the 4 receiving-sets selected by the engineer are defective.

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Answer from: lucaszsanches

0.0430 = 4.30% probability that exactly 2 of the 4 receiving-sets selected by the engineer are defective.

Step-by-step explanation:

Sets are chosen from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

Lots of 50 means that N = 50

5 are defective, which means that k = 5

4 are selected, which means that n = 4

Find the probability that exactly 2 of the 4 receiving-sets selected by the engineer are defective.

This is P(X = 2) . So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 2) = h(2,50,4,5) = \frac{C_{5,2}*C_{45,2}}{C_{50,4}} = 0.0430$

0.0430 = 4.30% probability that exactly 2 of the 4 receiving-sets selected by the engineer are defective.

Answer from: Quest

The full question isn’t shown. part of it is cut off
Ihave a test on this i need an in depth explanation i tried to solve this multiple ways

Answer from: Quest

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.

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