Each problem is a complete scenario. Solutions show: Model notation (e.g., X ~ N(μ, σ²)), Given, Find, Sketch (verbally), Compute (with z), and Excel formulas.
A student buys a new iPhone. The battery life on a full charge is modeled as normal with mean 12 hours and standard deviation 2 hours. What is the probability that a randomly chosen full charge lasts at most 14 hours? (Round probability to 4 decimals.)
Model \(X \sim \mathcal{N}(\mu=12,\,\sigma^2=2^2)\)
Given: \(\mu=12\) h, \(\sigma=2\) h. Find: \(P(X\le 14)\).
Sketch: Normal curve centered at 12; shade left of 14.
Compute: \(z=\dfrac{14-12}{2}=1\Rightarrow P=\Phi(1)=\mathbf{0.8413}\).
=NORM.DIST(14, 12, 2, TRUE) → 0.8413447
ENGR200 midterm scores are normal with mean 78 and sd 10. What is the probability a randomly chosen student scores above 90? (Round to 4 decimals.)
Model \(X \sim \mathcal{N}(78,\,10^2)\)
Find: \(P(X>90)=1-\Phi\!\big((90-78)/10\big)\).
Compute: \(z=1.2\Rightarrow P=1-\Phi(1.2)=\mathbf{0.1151}\).
=1 - NORM.DIST(90, 78, 10, TRUE)
Diameters of machined shafts are \(N(10.00, 0.05^2)\) mm. Parts are accepted if 9.95 ≤ X ≤ 10.05 mm. What proportion pass inspection? (Round to 4 decimals.)
Model \(X \sim \mathcal{N}(10.00,\,0.05^2)\)
Find: \(P(9.95\le X\le 10.05)=\Phi(1)-\Phi(-1)\).
Compute: \(z_1=-1,\ z_2=1\Rightarrow P=0.8413-0.1587=\mathbf{0.6827}\).
=NORM.DIST(10.05,10,0.05,TRUE) - NORM.DIST(9.95,10,0.05,TRUE)
Find the z-value that leaves 2.5% in the upper tail of the standard normal distribution.
Model \(Z \sim \mathcal{N}(0,1)\)
Find: \(z_{0.975}=\Phi^{-1}(0.975)=\mathbf{1.95996}\approx 1.96\).
=NORM.S.INV(0.975)
A student’s commute time is \(N(28,6^2)\) minutes. What commute time is the 80th percentile (so 80% of commutes take ≤ this long)? (Round to 2 decimals.)
Model \(X \sim \mathcal{N}(28,\,6^2)\)
Find: \(x_{0.80}=\mu+\sigma\,z_{0.80}\) with \(z_{0.80}=0.8416\).
Compute: \(x=28+6(0.8416)=\mathbf{33.05}\) min.
=NORM.INV(0.80, 28, 6)
An electronics lab models LED lifetime as \(N(50{,}000, 8{,}000^2)\) hours. Find the central 90% interval \([L,U]\) (5% in each tail). Round to the nearest hour.
Model \(X \sim \mathcal{N}(50\,000,\,8\,000^2)\)
Find: \(L=\mu-1.6449\,\sigma\), \(U=\mu+1.6449\,\sigma\).
Compute: \(L=50{,}000-1.6449(8{,}000)=\mathbf{36{,}841}\); \(U=50{,}000+1.6449(8{,}000)=\mathbf{63{,}159}\).
=NORM.INV(0.05, 50000, 8000) and =NORM.INV(0.95, 50000, 8000)
Course averages are \(N(75,12^2)\). Find the cutoff score \(x\) such that only 10% of students score above \(x\). (Round to 2 decimals.)
Model \(X \sim \mathcal{N}(75,\,12^2)\)
Find: \(x=\mu+\sigma\,z\) with lower-tail 0.90 → \(z=1.2816\).
Compute: \(x=75+12(1.2816)=\mathbf{90.38}\).
=NORM.INV(0.90, 75, 12)
A thermistor’s temperature reading is \(N(100,5^2)\) °C. What is the z-score for 93 °C, and what is \(P(X<93)\)? (Round probability to 4 decimals.)
Model \(X \sim \mathcal{N}(100,\,5^2)\)
Compute z: \(z=(93-100)/5=\mathbf{-1.4}\).
Find probability: \(P=\Phi(-1.4)=\mathbf{0.0808}\).
=NORM.DIST(93, 100, 5, TRUE)
Bolt length is \(N(50,0.2^2)\) mm. Bolts are acceptable if 49.7 ≤ X ≤ 50.3 mm. Find the total defect probability \(P(X<49.7)+P(X>50.3)\). (Round to 4 decimals.)
Model \(X \sim \mathcal{N}(50,\,0.2^2)\)
Compute: \(z_L=(49.7-50)/0.2=-1.5\Rightarrow P_L=\Phi(-1.5)=0.0668\). \(z_U=(50.3-50)/0.2=1.5\Rightarrow P_U=1-\Phi(1.5)=0.0668\).
Answer: Total defect \(=0.0668+0.0668=\mathbf{0.1336}\).
=NORM.DIST(49.7, 50, 0.2, TRUE) + (1 - NORM.DIST(50.3, 50, 0.2, TRUE))
Suppose \(X\sim \mathcal{N}(\mu=70,\,\sigma^2)\). If only 5% of values exceed 85, find \(\sigma\). (Round to 2 decimals.)
Model \(X \sim \mathcal{N}(70,\,\sigma^2)\)
Translate tail: \(P(X>85)=0.05 \Rightarrow P(X\le 85)=0.95\Rightarrow (85-70)/\sigma=z_{0.95}=1.6449\).
Compute: \(\sigma=15/1.6449=\mathbf{9.12}\).
=15 / NORM.S.INV(0.95)