Róndelo draws the semicircle shown at the right what is the area of the semicircle

For a normal distribution with mean 47 and standard deviation 6, find the probability of obtaining a value less than 45 or great

Llana [10]

Answer:

0.7392 = 73.92% probability of obtaining a value less than 45 or greater than 49.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean 47 and standard deviation 6

This means that $\mu = 47, \sigma = 6$

Less than 45:

p-value of Z when X = 45, so:

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

$Z = \frac{45 - 47}{6}$

$Z = -0.3333$

$Z = -0.3333$ has a p-value of 0.3696.

More than 49:

1 subtracted by the p-value of Z when X = 49. So

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

$Z = \frac{49 - 47}{6}$

$Z = 0.3333$

$Z = 0.3333$ has a p-value of 0.6304.

1 - 0.6304 = 0.3996

Less than 45 or greater than 49:

2*0.3696 = 0.7392

0.7392 = 73.92% probability of obtaining a value less than 45 or greater than 49.

4 0

3 years ago

Find two consecutive even positive integers whose product is 624.

dimaraw [331]

Let x=first even number then x+2 would be the next even number. (x)(x+2)=624. x^2 +2x=624. x^2 +2x -624=0. Factor or use the quadratic formula. (x+26)(x-24)=0 x=-26 or x=24 We want the positive solution so x=24 and x+2=26 24*26=624 so it works.
A square root can start us ...
625 = 25 x 25 ==> 24 x 26 will do it!

4 0

4 years ago

A parking lot contains 100 cars , k of which happen to be lemons. we select m of these cars

Sunny_sXe [5.5K]

Given that a parking lot contains 100 cars, k of which happen to be lemons.

This is a conditional probability question.

Let event A be that a car is tested and event B be that a car is lemon.

The probability that a car is lemon is given by $\frac{k}{100}$

The probability that a car is tested is given by $\frac{m}{100}$

The probability that a car is lemon and it is tested is given by $\frac{n}{m}$

For a conditional probability, the probablility of event A given event B is given by:

$P(A|B)= \frac{P(A\cap B)}{P(B)}$

Therefore, the probability that a car is lemon, given that it is tested is given by.

$P(lemon|tested)= \frac{P(lemon\, and\, tested)}{P(tested)} \\ \\ = \frac{ \frac{n}{m} }{ \frac{m}{100} } = \frac{n}{m} \times \frac{100}{m} = \frac{100n}{m^2}$