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erastovalidia [21]
3 years ago
14

A local bank surveyed the status of ASU student accounts and found that the average overdraft was $21.22 with a standard deviati

on of $5.49. If the distribution is normal, find the probability of a student being overdrawn by more than $18.75.
Mathematics
1 answer:
svlad2 [7]3 years ago
4 0

Answer: the probability of a student being overdrawn by more than $18.75 is 0.674

Step-by-step explanation:

Since the bank overdrafts of ASU student accounts are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = bank overdraft of Asu students.

µ = mean

σ = standard deviation

From the information given,

µ = $21.22

σ = $5.49

We want to find the probability of a student being overdrawn by more than $18.75. It is expressed as

P(x > 18.75) = 1 - P(x ≤ 18.75)

For x = 18.75,

z = (18.75 - 21.22)/5.49 = - 0.45

Looking at the normal distribution table, the probability corresponding to the z score is 0.326

Therefore,

P(x > 18.75) = 1 - 0.326 = 0.674

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<u>ANSWER:  </u>

The length and breadth of the rectangle are 18 m and 12 m.

<u>SOLUTION: </u>

Let the length and breadth of a rectangle be "l" and "b"

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\begin{array}{l}{\frac{l}{b}=\frac{3}{2}} \\\\ {l=\frac{3 b}{2}}\end{array} -- eqn 1

After changing the length and breadth by 1 meter on both sides, length and breadth becomes L+2 and b+2

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\frac{l+2}{b+2}=\frac{10}{7}

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7l + 14 = 10b + 20

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Now, substitute “l” value in (2)

10 b-7\left(\frac{3 b}{2}\right)+6=0

10 b-\frac{21 b}{2}+6=0

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