study-time norm. dist., 22% of 75 spend 47 min; Aaron spent

[juicyyflavored]

The time AP students spend on hw each evening is normally distributed. 22% of 75 students spend 47 minutes doing hw. Aaron spent 65 on hw and had a standarized score of 1.932. Find the mean and the standard deviation. Find the approx number of students who spend less than 20 min. on hw?

 

[soroban]

Re: system of equations/ statistics

Hello, juicyyflavored!

$\displaystyle \text{We're expected to know this formula: }\;z \;=\;\frac{x - \mu}{\sigma} \;\;\text{where: }\:\begin{Bmatrix}x &=& \text{raw score} \\ \mu &=& \text{mean} \\ \sigma &=& \text{std.dev.} \end{Bmatrix}$ .[1]

The time AP students spend on hw each evening is normally distributed.
22% of the 75 students spend 47 minutes or less doing hw.
Aaron spent 65 on hw and had a standarized score of 1.932.
(a) Find the mean and the standard deviation.

$\displaystyle \text{Aaron had: }\:x = 65,\;z = 1.932$
. . $\displaystyle \text{Substitute into [1]: }\:1.932 \:=\:\frac{65 - \mu}{\sigma} \quad\Rightarrow\quad \mu + 1.932\sigma \:=\:65$ .[2]


$\displaystyle \text{The class had 22 percent below 47}$.
$\displaystyle \text{This corresponds to: }\:z = \text{-}0.77$
. . \(\displaystyle \text{Substitute into [1]: }\:\text{-}0.77 \:=\:\frac{47 - \mu}{\sigma} \qyad\Rightarrow\quad \mu - 0.77\sigma \:=\:47\) .[3]

$\displaystyle \text{Subtract [3] from [2]: }\;2.702\sigma \:=\:18 \quad\Rightarrow\quad\boxed{ \sigma \:\approx\:6.66}$

$\displaystyle \text{Substitute into [1]: }\:\mu + 1.932(6.66) \:=\:65 \quad\Rightarrow\quad\boxed{\mu \:\approx\:52.13}$

(b) Find the approx number of students who spend less than 20 min. on hw?

$\displaystyle \text{We have: }\:x = 20$

$\displaystyle \text{Then: }\:z \:=\:\frac{20-52.13}{6.66} \:=\:-4.82$


$\displaystyle \text{The area to left of }z = -4.82\text{ is less than 0.002.}$

. . $\displaystyle \text{Hence: }\:0.002 \times 75 \:=\:0.15 \:\approx\:0$

$\displaystyle \text{Therefore, there are }no\text{ students who spend less than 20 minutes on hw.}$