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BARSIC [14]
2 years ago
15

The ratio of boys to girls in a group is 5:3. If there are 18 more boys than girls, work out how

Mathematics
1 answer:
zmey [24]2 years ago
8 0
<h3>Answer:  72 people total</h3>

45 boys, 27 girls

============================================================

Explanation:

g = number of girls

g+18 = number of boys, since there are 18 more boys than girls

For example, if g = 7, then g+18 = 7+18 = 25. In this example, we have 7 girls and 25 boys. The problem with this example is the ratio of boys to girls is 25:7 which is not equal to 5:3

We could use trial and error until we land on the solution, or we can use algebra as shown in the next section.

-------------------

The ratio of boys to girls is 5:3, meaning

(number of boys)/(number of girls) = 5/3

which turns into

(g+18)/g = 5/3

after we make the proper substitutions

Let's solve for g

(g+18)/g = 5/3

3(g+18) = 5g

3g+54 = 5g

54 = 5g-3g

54 = 2g

2g = 54

g = 54/2

g = 27

There are 27 girls.

This leads to,

g+18 = 27+18 = 45

So there are 45 boys

In total, we have 27+45 = 72 people

-----------------------------

Checking the answer:

The claim is that 45/27 is the same as 5/3

For now assume that claim is true. Set the fractions equal to each other and use cross multiplication like so

45/27 = 5/3

45*3 = 27*5

135 = 135

The third equation has the same thing on both sides, so it's a true equation. This must mean the first equation is true as well. Therefore, 45:27 and 5:3 are the same ratio and the answer is confirmed.

You could also use your calculator to note that

45/27 = 1.667

5/3 = 1.667

both decimal values are approximate.

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Answer:

P(T>8300000)=1-P(T

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Data given

n = 25000 represent the automobile policy holders

\mu= 320 represent the population mean

\sigma =540 represent the population standard deviation

Let T the variable that represent the total of interest on this case. We can assume that the random variable for an individual policy holder is given by:

X\sim N(\mu = 540, \sigma=540)

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Solution to the problem

First we need to find the distribution for the random variable T like this:

\bar X = \frac{\sum_{i=1}^n x_i}{n}

And the total T is given by:

T=\sum_{i=1}^n X_i =n \bar X

We can find the expected value, variance and deviation for this random variable like this:

E(T)= n E(\bar X) = n \mu = 25000*320=8000000

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Sd(T)=\sqrt{7290000000}=85381.497

And we are interested on this probability:

P(T>8300000)

And we can use the Z score formula given by:

Z=\frac{T-E(T)}{\sigma_T}

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