Given : The incomes of families in Newport Harbor are normally distributed with Mean : $\mu=\$ 750,000$ and Standard deviation : $\sigma= $250,000$

Samples size : n=4

Let x be the random variable that represents the incomes of families in Newport Harbor.

The z-statistic :-

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For x= $800,000

$z=\dfrac{800000-750000}{\dfrac{250000}{\sqrt{4}}}=0.4$

By using the standard normal distribution table , we have

The probability that the average income of these 4 families exceeds $800,000 :-

$P(x>80,000)=P(z>0.4)=1-P(z\leq0.4)\\\\=1-0.6554217=0.3445783\approx0.345$

Hence, the probability that the average income of these 4 families exceeds $800,000 =0.345

6 0

4 years ago

I need to know the answer please help me

spin [16.1K]

First is 42 and the second is 4.5

6 0

2 years ago

A savings bank invests $58,800 in municipal bonds and earns 12% per year on the investment. How much money is earned per year?

seraphim [82]

To find the answer we simply have to find 12% of 58,800. So to do that, we can multiply it by .12

58,800 • .12 = 7,056

So $7,056 is earned per year

5 0

4 years ago

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