Question

AP STATISTIC

Answer 1

The mean for the total cost of the two items is 82. The standard deviation of the total cost of the two items is 14.14214. The probability of finding two random items at this auction with a total price of less than $80 is 0.44377.

What is a random variable?

A random variable is a variable with an undetermined value that gives values to each of the results of a statistical experiment.

From the parameters given:

• Let us assume that X represents the random variable that connotes the price of the item during the large auction.

Given that:

• X is normally distributed with a mean of $41 and
• A standard deviation of $10

X $\sim$ N(μ, σ²)

X $\sim$ N(41, 10²)

Suppose we made an assumption that Y should denote the total cost of items:

i.e.

Y = X₁ + X₂

Here;

$\mathbf{X_i \sim N(41, 10^2)} \\ \\ \mathbf{E(Y) = E(X_1+X_2) } \\ \\ \mathbf{E(Y) = 41 + 41 } \\ \\ \mathbf{E(Y) = 82}$

The variance of (Y) is:

$\mathbf{Var (Y) = Var(X_1+X_2) = Var (X_1) + Var(X_2) } \\ \\\mathbf{= 10^2+ 10^2} \\ \\ \mathbf{=200}$

$\mathbf{Standard \ deviation \ SD (Y) = \sqrt{200} }$

= 14.14214

The probability of finding the two random items at the auction with a total price of less than $80 can be computed as:

P(Y < 80)  

Since the data is normally distributed,

$\mathbf{=P\Big(Z < \dfrac{x -\mu}{\sigma}\Big)}$

$\mathbf{=P\Big(Z < \dfrac{80 -82}{14.14214}\Big)}$

$\mathbf{=P\Big(Z < -0.1414213\Big)}$

Recall that:

P(Z < -z) = P(Z > z)

Hence;

= P (Z  > 0.1414213)

= 1 - P(Z ≤  0.1414213)

From the Z tables, the value of Z at 0.1414213 is 0.55623;

=  1 -  0.55623

= 0.44377

Therefore, we can conclude that the probability of finding two random items at this auction with a total price of less than $80 is 0.44377.

Learn more about random variables in probability here:

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