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igomit [66]
3 years ago
9

The average amount of money spent for lunch per person in the college cafeteria is $6.35 and the standard deviation is $2.31. Su

ppose that 41 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 4 decimal places where possible.
What is the distribution of X?X ~ N(,)
What is the distribution of x¯? x¯ ~ N(,)
For a single randomly selected lunch patron, find the probability that this patron's lunch cost is between $6.1605 and $6.757.
For the group of 17 patrons, find the probability that the average lunch cost is between $6.1605 and $6.757.
Mathematics
1 answer:
alexdok [17]3 years ago
8 0

Answer:

(6.35, 5.34)

(6.35, 0.99)

0.10253

0.3984

Step-by-step explanation:

Given that :

μ = 6.35

σ = 2.31

n = 41

What is the distribution of X?X ~ N(,)

X :

Mean of distribution = μ= 6.35

The variance = σ^2 = 2.31^2 = 5.3361 = 5.34

X ~ N = (6.35, 5.34)

What is the distribution of x¯? x¯ ~ N(,)

The mean = μ = 6.35

The standard deviation of the mean = σ/sqrt(n) = 6.35/sqrt(41) = 0.9917 = 0.99

X ~N = (6.35, 0.99)

find the probability that this patron's lunch cost is between $6.1605 and $6.757.

P(6.1605 < x < 6.757)

Obtain the standardized scores:

Z = (x - μ) / σ ; (6.1605 - 6.35) / 2.31 = - 0.082

P(Z < - 0.082) = 0.46732 (Z probability calculator)

(6.757 - 6.35) / 2.31 = 0.176

P(Z < 0.176) = 0.56985 (Z probability calculator)

P(Z < 0.176) -P(Z < - 0.082)

0.56985 - 0.46732 = 0.10253

For the group of 17 patrons, find the probability that the average lunch cost is between $6.1605 and $6.757.

P(6.1605 < x < 6.757) ; n = 17

Obtain the standardized scores:

Z = (x - μ) / σ/sqrt(n) ; (6.1605 - 6.35) / 2.31/sqrt(17) = - 0.338

P(Z < - 0.338) = 0.36768 (Z probability calculator)

(6.757 - 6.35) / 2.31/sqrt(17) = 0.726

P(Z < 0.726) = 0.76608 (Z probability calculator)

P(Z < 0.726) -P(Z < - 0.338)

0.76608 - 0.36768 = 0.3984

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Answer:

  • a) 3/5·((-2)^n + 4·3^n)
  • b) 3·2^n - 5^n
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  • e) 2 + 3·(-1)^n
  • f) (-3)^n·(3 - 2n)
  • g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form a[n]=r^n, then it will satisfy ...

  r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}

Rearranging and dividing by r^{n-2}, we get the quadratic ...

  r^2-c_1r-c_2=0

The quadratic formula tells us values of r that satisfy this are ...

  r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

  a[n]=pr_1^n+qr_2^n

We can find p and q by solving the initial condition equations:

\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]

These have the solution ...

p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}

_____

Using these formulas on the first recurrence relation, we get ...

a)

c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n

__

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

  a[n]=(p+qn)r^n

The initial condition equations are now ...

\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]

and the solutions for p and q are ...

p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}

__

Using these formulas on problem (d), we get ...

d)

c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n

__

And for problem (f), we get ...

f)

c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

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3 years ago
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15 different number, yah you ar right
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