Every solution shows the conditions, the formula and the numbers, so you can find the exact step where your work differs. Normal areas use Table A values for z-scores rounded to two decimals; a calculator may differ in the last digit.
Problem 1: Coin flips
Check BINS. Each flip is binary (heads or tails). Flips are independent. The number of flips is fixed at \(12\). The chance of heads is \(0.5\) on every flip.
All four conditions hold, so \(X\) is binomial with \(n=12\) and \(p=0.5\).
Problem 2: Cards without replacement
The trials are not independent. On the first card the chance of a heart is \(\frac{13}{52}=0.25\). If the first card is a heart, the second is a heart with probability \(\frac{12}{51}\approx 0.2353\); otherwise it is \(\frac{13}{51}\approx 0.2549\). The sample of \(10\) is also more than \(10\%\) of the \(52\) cards, so the 10% rule does not rescue independence.
Not binomial: the trials are dependent and the success probability changes.
Problem 3: Rolling until a six
A binomial count needs a fixed number of trials \(n\) chosen in advance. Here Ethan stops when the first \(6\) appears, so the number of rolls varies from one run to the next.
Not binomial: \(n\) is not fixed.
Problem 4: Exactly three successes
Use \(P(X=k)={_{n}C_{k}p^{k}(1-p)^{n-k}\) with \(k=3\).
\(P(X=3)={_{8}C_{3}(0.25)^{3}(0.75)^{5}=56(0.015625)(0.2373)\approx\) 0.2076
Calculator check: binompdf(8, 0.25, 3).
Problem 5: No successes
Here \(k=0\) and \({_{6}C_{0}=1\), so only the failure factor remains.
\(P(X=0)={_{6}C_{0}(0.1)^{0}(0.9)^{6}=(0.9)^{6}=0.531441\approx\) 0.5314
Problem 6: At most two
\(P(X\le 2)=P(0)+P(1)+P(2)\).
- \(P(0)=(0.6)^{5}=0.07776\)
- \(P(1)=5(0.4)(0.6)^{4}=0.2592\)
- \(P(2)=10(0.4)^{2}(0.6)^{3}=0.3456\)
Sum: \(0.07776+0.2592+0.3456=0.68256\). \(P(X\le 2)\approx 0.6826\) (binomcdf(5, 0.4, 2)).
Problem 7: At least one
Use the complement: \(P(X\ge 1)=1-P(X=0)\).
\(P(X=0)=(0.95)^{10}\approx 0.5987\), so \(P(X\ge 1)=1-0.5987=\) 0.4013.
Problem 8: Mean and standard deviation
\(\mu=np=40(0.35)=14\).
\(\sigma=\sqrt{np(1-p)}=\sqrt{40(0.35)(0.65)}=\sqrt{9.1}\approx 3.017\).
Mean \(14\), standard deviation \(\approx 3.017\).
Problem 9: Interpreting the mean
\(\mu=200(0.02)=4\) and \(\sigma=\sqrt{200(0.02)(0.98)}=\sqrt{3.92}\approx 1.980\).
Interpretation: in the long run, shipments of \(200\) cases average \(4\) cracked cases per shipment. Individual shipments vary, typically by about \(2\) cases from \(4\).
Problem 10: Reading a distribution graph
The shaded bars are \(X=3\) and \(X=4\): \(0.25+0.0625=0.3125\).
Check: \({_{4}C_{3}(0.5)^{4}=\frac{4}{16}\) and \({_{4}C_{4}(0.5)^{4}=\frac{1}{16}\), so \(\frac{5}{16}=0.3125\). \(P(X\ge 3)=0.3125\)
Problem 11: Probability of a middle range
Each outcome has probability \(\frac{{_{8}C_{k}}{256}\). The counts are \({_{8}C_{3}=56\), \({_{8}C_{4}=70\) and \({_{8}C_{5}=56\).
\(P(3\le X\le 5)=\frac{56+70+56}{256}=\frac{182}{256}\approx\) 0.7109
Problem 12: z-score of a height
\(z=\frac{75-69}{3}=2.00\).
He is \(2\) standard deviations above the mean height. By the 68-95-99.7 rule, only about \(2.5\%\) of men are taller.
Problem 13: Area to the right
\(z=\frac{620-500}{100}=1.20\). Table A gives an area to the left of \(0.8849\).
\(P(X> 620)=1-0.8849=\) 0.1151, so about \(11.5\%\) of students score above \(620\).
Problem 14: Area between two values
Lower: \(z=\frac{65-72}{6}=-1.17\), area \(0.1210\). Upper: \(z=\frac{80-72}{6}=1.33\), area \(0.9082\).
\(P(65< X< 80)=0.9082-0.1210=\) 0.7872.
Problem 15: Finding a percentile
The 90th percentile has area \(0.90\) to its left. In Table A the closest area, \(0.8997\), goes with \(z=1.28\).
\(x=\mu+z\sigma=100+1.28(15)=100+19.2=\) 119.2.
Problem 16: Empirical rule
(a) \(15.6\) and \(16.4\) are \(2\) standard deviations from \(16.0\) (\(16.0\pm 2(0.2)\)). About 95% of bags.
(b) \(15.8\) is \(1\) standard deviation below the mean. About \(68\%\) lie within \(1\) SD, leaving \(32\%\) in the two tails, so \(16\%\) lie below. About 16% of bags.
Problem 17: Comparing z-scores
Maya: \(z=\frac{84-75}{6}=1.50\). Diego: \(z=\frac{88-80}{8}=1.00\).
Maya did better relative to her class: she is \(1.5\) SDs above the mean, Diego only \(1.0\).
Problem 18: Checking the approximation
(a) \(np=150(0.2)=30\ge 10\) and \(n(1-p)=150(0.8)=120\ge 10\). Reasonable.
(b) \(np=40(0.1)=4\), which is less than \(10\) (and \(n(1-p)=36\)). Not reasonable, because the distribution is skewed right.
Problem 19: Normal approximation
(a) \(np=30\ge 10\) and \(n(1-p)=70\ge 10\). The approximation is reasonable.
(b) \(\mu=30\), \(\sigma=\sqrt{100(0.3)(0.7)}=\sqrt{21}\approx 4.583\). \(z=\frac{35-30}{4.583}=1.09\), and \(P(Z\ge 1.09)=1-0.8621=\) 0.1379.
(c) The bar for \(X=35\) extends from \(34.5\) to \(35.5\), but the normal area starts at exactly \(35\), so it misses half of that bar. The approximation \(0.1379\) is therefore smaller than the exact \(0.1629\).
Problem 20: Free throws
Conditions: binary, independent, \(n=15\), \(p=0.7\).
\(P(X\ge 12)=1-P(X\le 11)=1-\text{binomcdf}(15, 0.7, 11)\).
\(P(12)+P(13)+P(14)+P(15)=0.2969\).
\(P(X\ge 12)\approx 0.2969\), so she makes at least 12 of 15 only about 30% of the time.
Problem 21: Overbooked flight
\(X\sim\) binomial, \(n=54\), \(p=0.9\). \(\mu=54(0.9)=48.6\); \(\sigma=\sqrt{54(0.9)(0.1)}=\sqrt{4.86}\approx 2.205\).
More than \(50\) show up means \(X\ge 51\): \(P(X\ge 51)=1-\text{binomcdf}(54, 0.9, 50)\approx\) 0.1985.
About 19.8% of flights like this would leave at least one passenger without a seat.
Problem 22: Recovering n and p
\(\mu=np=24\) and \(\sigma^{2}=np(1-p)=9.6\).
Divide: \(1-p=\frac{9.6}{24}=0.4\), so \(p=0.6\).
Then \(n=\frac{24}{0.6}=40\). \(n=40\), \(p=0.6\). Check: \(40(0.6)(0.4)=9.6\).
Problem 23: Poll approximation
Conditions: \(np=200(0.45)=90\ge 10\) and \(n(1-p)=110\ge 10\).
\(\mu=90\), \(\sigma=\sqrt{200(0.45)(0.55)}=\sqrt{49.5}\approx 7.036\). \(z=\frac{100-90}{7.036}=1.42\).
\(P(X> 100)\approx 1-0.9222=\) 0.0778. (The exact binomial value is about 0.0681.)
Problem 24: Is the claim believable?
(a) \(\mu=150(0.03)=4.5\); \(\sigma=\sqrt{150(0.03)(0.97)}=\sqrt{4.365}\approx 2.089\).
(b) No. \(np=4.5< 10\), so the distribution is skewed and the normal approximation is not appropriate.
(c) Use the exact binomial: \(P(X\ge 10)=1-\text{binomcdf}(150, 0.03, 9)\approx\) 0.0155.
If the claim were true, a result this extreme would occur only about 1.6% of the time. That is rare enough to cast real doubt on the claim.
See every chapter on the AP Statistics math page.