Question

100 POINTS PLEASE HELP!

The scores on a standardized exam are normally distributed with a mean of 245 and a standard deviation of 54.
Use technology or a z-score table to answer the questions.
(a) Bailey scored higher than 90% of all other people taking the exam. What was Bailey's approximate score? Round to the nearest whole number.
(b) Approximately what percent of people taking the exam scored between 200 and 300? Round to the nearest whole percent.

Answer 1

a) 314

b) 64%

Explanation:

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

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

Given:

• Mean μ = 245
• Standard deviation σ = 54

Therefore, if the scores on a standardized exam are normally distributed:

`\boxed{X \sim\text{N}(245,54^2)}`

where X is the score on the exam.

Converting to the Z distribution

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

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%:

`\implies \text{P}(X > a) =0.9`

Method 1

Transform X to Z:

`\text{P}(X > a) = \text{P}\left(Z > (a-245)/(54)\right)=0.9`

If using z-tables:

`\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}`

According to the z-tables, when p = 0.1000, z = 1.2816

`\implies (a-245)/(54)=1.2816`

`\implies a-245=69.2064`

`\implies a=314.2064`

Method 2

Calculator input for "inverse normal":

• Area = 0.9
• μ = 245
• σ = 54

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.