Answer:
a) 314
b) 64%
Step-by-step explanation:
If a continuous random variable X is normally distributed with mean μ and variance σ², it is written as:
[tex]\boxed{X \sim\text{N}(\mu,\sigma^2)}[/tex]
Given:
- Mean μ = 245
- Standard deviation σ = 54
Therefore, if the scores on a standardized exam are normally distributed:
[tex]\boxed{X \sim\text{N}(245,54^2)}[/tex]
where X is the score on the exam.
Converting to the Z distribution
[tex]\boxed{\textsf{If }\: X \sim\textsf{N}(\mu,\sigma^2)\:\textsf{ then }\: \dfrac{X-\mu}{\sigma}=Z, \quad \textsf{where }\: Z \sim \textsf{N}(0,1)}[/tex]
Part (a)
If Bailey scores higher than 90% of all the other people taking the exam, to calculate his score, we need to find the value of a for which P(X > a) = 90%:
[tex]\implies \text{P}(X > a) =0.9[/tex]
Method 1
Transform X to Z:
[tex]\text{P}(X > a) = \text{P}\left(Z > \dfrac{a-245}{54}\right)=0.9[/tex]
If using z-tables:
[tex]\begin{aligned}\text{P}(X > a) =1 - \text{P}(X \leq a) &= 0.9 \\\text{P}(X \leq a)&=1-0.9\\ \text{P}(X \leq a)&=0.1\end{aligned}[/tex]
According to the z-tables, when p = 0.1000, z = 1.2816
[tex]\implies \dfrac{a-245}{54}=1.2816[/tex]
[tex]\implies a-245=69.2064[/tex]
[tex]\implies a=314.2064[/tex]
Method 2
Calculator input for "inverse normal":
xInv = 314.2037885...
Therefore, Bailey's approximate score was 314 (nearest whole number).
Part (b)
To calculate the percent of people taking the exam who scored between 200 and 300, we need to find P(200 < X < 300).
Calculator input for "normal cumulative distribution function (cdf)":
- Upper bound: x = 300
- Lower bound: x = 200
- μ = 245
- σ = 54
⇒ P = 0.6434558166...
⇒ P = 64.34558166...%
Therefore, approximately 64% (nearest whole percent) of people taking the exam scored between 200 and 300.
