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Hey!
To solve this equation, we must first convert the mixed numbers into improper fractions. To do that we'll multiply the denominator by the whole number and add the total to the numerator.
<em>Original Equation :</em>
<em>New Equation {Converted Mixed Numbers Into Improper Fractions} :</em>
Now that we've successfully changed our equation, we'll move onto the next step. Next we'll have change the fractions one more time so that the denominators are the same. To do that we'll multiply two by eight to get sixteen. We'll also be multiplying the top portion of the fraction by eight, because whatever you do to the bottom must also be done to the top; And vise versa.
<em>Fraction Conversion :</em>
So, now that we have our new fraction we can go ahead and plug that into our equation so we can start adding. Remember, we don't add the denominators, just the numerators.
<em>Old Equation :</em>
<em>New Equation {Input New Fraction With Common Denominator} :</em>
<em>Solution {New Equation Solved} :</em>
Now I'm not exactly sure how your instructor will accept your answer, but most prefer the final answer to be in the simplest form. In other words, they don't want it in the form of an improper fraction. They'd most likely rather have it as a mixed number.
<em>New Solution {Converted to a Mixed Number} :</em>
Since the fraction cannot be simplified anymore, this should be our final answer.
<em>So, this means that the answer to the equation is </em> .
Hope this helps!
- Lindsey Frazier ♥
Answer:
μ=23.710740....
σ=3.592592....
Step-by-step explanation:
At first, we are told that it´s a normal distribution and the following:
P(x≤15.34)=0.1
P(x≤16.31)=0.2
We start looking for those values (0.1 and 0.2) and the probabilities that gives us those values in a normal standard distribution. There´s some tools that´ll help us out but i´ll be using a chart with the values of some probabilities in a normal standard distribution (That it´s attached).
The probability of P(z≤ a.b) where "a" is the whole part of a number and "b" the decimal part is in the coordinates (a,b). For example for our problem:
P(z≤-2.33)=0.099≈0.1
P(z≤-2.06)=0.197≈0.2
To find a probability of a normal (not standard) distribution we use a method called Normalize that proceeds:
P(x≤a)=P( (x-μ)/σ ≤ (a-μ)/σ )
Where μ is the Mean of the data and σ is the Standard deviation. P( (x-μ)/σ ≤ (a-μ)/σ ) is now part of a normal standard distribution and we are able to look for it in the chart.
We will use it to find the Mean and the STD for our distribution.
P(x≤15.34)=P(z≤-2.33)≈0.1
P(x≤16.31)=P(z≤-2.06)≈0.2
so we have the equations:
(15.34 - μ)/σ= -2.33 and (16.31 - μ)/σ= -2.06
And we solve them for μ and σ:
(15.34 - μ)/-2.33= σ and (16.31 - μ)/-2.06= σ
(15.34 - μ)/-2.33 = (16.31 - μ)/-2.06
2.06*(15.34 - μ) = 2.33(16.31 - μ)
2.33μ - 2.06μ = 2.33(16.31) - 2.06(15.34)
μ= 6.4019 / 0.27
μ= 23.710740...
With this value we find σ in one equation:
(15.34 - (23.710740...)/σ = -2.33
(15.34 - (23.710740...)/-2.33 = σ
σ=3.592592...
We finally have our answers:
μ=23.710740....
σ=3.592592....
