Question

Provide an appropriate response.

​Find the area of the shaded region. The graph depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test).

Answer 1

0.7938

Explanation:

The given graph is a normal distribution curve.

If a continuous random variable X is normally distributed with mean μ and variance σ²:

`\boxed{X \sim \textsf{N}(\mu,\sigma^2)}`

Given:

• mean μ = 100
• standard deviation σ = 15

`\text{If \; $X \sim \textsf{N}(100,15^2)$,\;\;find\;\;P$(85\leq X\leq 125)$\;\;to\;3\;s.f.}`

Therefore, we need to find the area to the left of x = 125 and subtract the area to the left of x = 85.

Method 1

Using a calculator:

`\begin{aligned}\implies \text{P}(85\leq X\leq 125)&= \text{P}(X\leq 125)-\text{P}(X < 85)\\&=0.9522096477-0.1586552539 \\&=0.7935543938\\&\approx0.7938\end{aligned}`

Method 2

Converting to the z-distribution.

`\boxed{\text{If\;\;$X \sim$N$(\mu,\sigma^2)$\;\;then\;\;$(X-\mu)/(\sigma)=Z$, \quad where $Z \sim$N$(0,1)$}}`

`x=85 \implies Z_1=(85-100)/(15)=-1`

`x=125 \implies Z_2=(125-100)/(15)=1.67`

Using the z-tables to find the corresponding probabilities (see attachments).

`\begin{aligned}\implies \text{P}(-1\leq Z\leq 1.67)&= \text{P}(Z\leq 1.67)-\text{P}(Z < -1)\\&=0.9525-0.1587\\&=0.7938\end{aligned}`

Answer 2

• A) 0.7938

---------------------------

Given

• μ = 100, σ = 15

Find the area between x = 85 and x = 125.

Find z-scores for each end

• z = (x - μ)/σ
• z₁ = (85 - 100)/15 = - 15/15 = - 1
• z₂ = (125 - 100)/15 = 25/15 ≈ 1.67

Find the corresponding probabilities from z-score table

• z₁ = 0.1587
• z₂ = 0.9525

Find the difference

• z₂ - z₁ = 0.9525 - 0.1587 = 0.7938

The matching choice is A