Second Midterm — Part 2 (Questions + Hidden Solutions)

Q1 • Coffee Dispenser Calibration (Normal)

Assume $X \sim \mathcal{N}(\mu=350, \sigma=12)$.

What proportion of cups are below 340 mL?
What proportion of cups are above 360 mL?
What proportion of cups are between 340 mL and 360 mL?
The café labels a cup “XL” if it is in the top 5% of volumes. What threshold (mL) triggers the label?
The cup’s capacity is 375 mL. What is $P(X>375)$?
On a day with 800 pours, how many overflows do you expect? Also report a 95% normal-approx range for the count.

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Model & method: standardize with $z=(x-\mu)/\sigma$; use $\Phi(z)$ for CDF.

(a) $z_{340}=-0.8333 \Rightarrow P=\Phi(z)=0.2023$.
(b) $z_{360}=0.8333 \Rightarrow P=1-\Phi(z)=0.2023$.
(c) $\Phi(0.8333)-\Phi(-0.8333)=0.5953$.
(d) $x_{0.95}=\mu+\sigma z_{0.95}=350+12(1.6449)=369.74\,\text{mL}$.
(e) $z_{375}=2.0833 \Rightarrow P=1-\Phi(z)=0.0186$.
(f) $E=800\times 0.0186=14.89$. 95% range: $7.40, 22.38$.

Excel: =NORM.DIST(340,350,12,TRUE), =1-NORM.DIST(360,350,12,TRUE), =NORM.DIST(360,350,12,TRUE)-NORM.DIST(340,350,12,TRUE), =NORM.INV(0.95,350,12), =1-NORM.DIST(375,350,12,TRUE). For counts: mean=800*p, sd=SQRT(800*p*(1-p)).

Q2 • Bins & Observed Counts (n=50)

X: T1=<5, T2=5–15, T3=15–30, T4=>30. Y: A=<$10, B=$10–20, C=$20–30, D=$30–40, E=>$40.

Y \\ X	T1<5	T2 5–15	T3 15–30	T4 >30	Row Sum
A: < $10	4	3	1	0	8
B: $10–20	2	5	3	1	11
C: $20–30	1	3	4	2	10
D: $30–40	0	1	4	5	10
E: > $40	0	0	3	8	11
Col Sum	7	12	15	16	50

Compute $f(X=T3, Y=D)$.
Compute $f_X(T3)$.
Compute $P(Y=E \mid X=T4)$.
Compute $P(X\ge15)$.
Compute $P(Y\ge30)$.
Check independence at (T4,E): compare $f(T4,E)$ with $f_X(T4)f_Y(E)$.

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Let $n=50$.
(1) $f(T3,D)=4/50=\mathbf{0.0800}$.
(2) $f_X(T3)=15/50=\mathbf{0.3000}$.
(3) $P(Y=E\mid X=T4)=\dfrac{8/50}{16/50}=\mathbf{0.5000}$.
(4) $P(X\ge15)=(15+16)/50=\mathbf{0.6200}$.
(5) $P(Y\ge30)=(10+11)/50=\mathbf{0.4200}$.
(6) Independence check: Left $=8/50=0.1600$. Right $=(16/50)(11/50)=0.0704$ ⇒ Not independent.

Q3 • Correlation & Variance (Study vs Sleep, 5 pairs)

Day	1	2	3	4	5
X (study)	3.5	4.0	5.0	6.0	6.5
Y (sleep)	7.8	7.2	6.5	5.9	5.4

Compute $\bar X$ and $\bar Y$.
Compute $s_X^2$ and $s_Y^2$ (sample variances).
Compute the sample covariance $s_{{XY}}$.
Compute Pearson $r=s_{{XY}}/(s_X s_Y)$.
Interpret the correlation (strength, direction, practical meaning).

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Means: $\bar X=5.000,\; \bar Y=6.560$.
Variances: $s_X^2=1.625,\; s_Y^2=0.933$.
Covariance: $s_{{XY}}=-1.225$.
Correlation: $r=-0.995$ (strong negative).
Interpretation: As study time increases, sleep decreases nearly linearly in this sample; near −1 indicates a very tight inverse association (not causation).