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1 year ago

(d) Given that n(§) = 96, n(A) = 50 and n(B) = 60. Find the maximum and minimum values for n(An B).

Mathematics

1 answer:

frez [133]1 year ago

3 0

Step-by-step explanation:

the max. value is when the smaller set (A) is completely contained in the larger set (B).

then n(A n B) is n(A) = 50.

the set intersection between A and B cannot get bigger than that. or A gets bigger ...

after all, the intersection means it is a set of all elements that exist in BOTH sets.

but then there must be other elements besides A and B in the universal set too, because n(universal set) = 96, and n(A u B) would be only 60.

the min. value could be the empty set or 0. but because n(universal set) = 96, and n(A) + n(B) = 110 and larger than 96, it means that there have to be some shared elements. at least 110 - 96 = 14 elements.

in this case there cannot be other elements in the universal set than A and B. and n(universal set) = n(AuB) = 96.

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A. Student GPAs: Bob’s z-score z = + 1.71 μ = 2.98 σ = 0.36 b. Weekly work hours: Sarah’s z-score z = + 1.18 μ = 21.6 σ = 7.1 c.

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Answer:

a) $x = \mu +z*\sigma$

And replacing we got:

$x= 2.98 + 1.71*0.36 = 3.5956$

b) $x = \mu +z*\sigma$

And replacing we got:

$x= 21.6 + 1.18*7.1 = 29.978$

c) $x = \mu -z*\sigma$

And replacing we got:

$x= 150 - 1.35*40= 96$

Step-by-step explanation:

Previous concepts

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 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".

Part a

Let X the random variable of interest for this case. We define the z score with the following formula:

$z=\frac{x-\mu}{\sigma}$

And for this case we know that $z = 1.71, \mu = 2.98,\sigma = 0.36$

If we solve for x from the z score formula we got:

$x = \mu +z*\sigma$

And replacing we got:

$x= 2.98 + 1.71*0.36 = 3.5956$

Part b

Let X the random variable of interest for this case. We define the z score with the following formula:

$z=\frac{x-\mu}{\sigma}$

And for this case we know that $z = 1.18, \mu = 21.6,\sigma = 7.1$

If we solve for x from the z score formula we got:

$x = \mu +z*\sigma$

And replacing we got:

$x= 21.6 + 1.18*7.1 = 29.978$

Part c

Let X the random variable of interest for this case. We define the z score with the following formula:

$z=\frac{x-\mu}{\sigma}$

And for this case we know that $z = -1.35, \mu = 150,\sigma = 40$

If we solve for x from the z score formula we got:

$x = \mu -z*\sigma$

And replacing we got:

$x= 150 - 1.35*40= 96$

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