The amount of money separating the lowest 80% of the amount invested from the highest 20% in a sampling distribution of 10 of the family's real estate holdings is $238,281.57.

Step-by-step explanation:

Let the random variable X represent the amount of money that the family has invested in different real estate properties.

The random variable X follows a Normal distribution with parameters μ = $225,000 and σ = $50,000.

It is provided that the family has invested in n = 10 different real estate properties.

Then the mean and standard deviation of amount of money that the family has invested in these 10 different real estate properties is:

$\mu_{\bar x}=\mu=\$225,000\\\\\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{50000}{\sqrt{10}}=15811.39$

Now the lowest 80% of the amount invested can be represented as follows:

$P(\bar X$

The value of z is 0.84.

*Use a z-table.

Compute the value of the mean amount invested as follows:

$\bar x=\mu_{\bar x}+z\cdot \sigma_{\bar x}$

$=225000+(0.84\times 15811.39)\\\\=225000+13281.5676\\\\=238281.5676\\\\\approx 238281.57$

Thus, the amount of money separating the lowest 80% of the amount invested from the highest 20% in a sampling distribution of 10 of the family's real estate holdings is $238,281.57.

6 0

3 years ago

Solve for w in the proportion.

Svet_ta [14]

Answer:

w=40

Step-by-step explanation:

5 0

3 years ago