Answer:
0.7938
Step-by-step explanation:
The given graph is a normal distribution curve.
If a continuous random variable X is normally distributed with mean μ and variance σ²:
[tex]\boxed{X \sim \textsf{N}(\mu,\sigma^2)}[/tex]
Given:
- mean μ = 100
- standard deviation σ = 15
[tex]\text{If \; $X \sim \textsf{N}(100,15^2)$,\;\;find\;\;P$(85\leq X\leq 125)$\;\;to\;3\;s.f.}[/tex]
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:
[tex]\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}[/tex]
Method 2
Converting to the z-distribution.
[tex]\boxed{\text{If\;\;$X \sim$N$(\mu,\sigma^2)$\;\;then\;\;$\dfrac{X-\mu}{\sigma}=Z$, \quad where $Z \sim$N$(0,1)$}}[/tex]
[tex]x=85 \implies Z_1=\dfrac{85-100}{15}=-1[/tex]
[tex]x=125 \implies Z_2=\dfrac{125-100}{15}=1.67[/tex]
Using the z-tables to find the corresponding probabilities (see attachments).
[tex]\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}[/tex]
