Probability Distributions · applications

Applications: Probability Distributions

10 question types · Pre-University · each worked step by step with a figure that follows the steps

H2

01

A Spinner Game at a School Fair, the Fee That Makes It Fair and the Top Prize That Leaves a Profit

methodList Every Prize with Its Probability, Weight Each Prize by Its Chance and Add; a Fair Fee Equals the Expected Prize

A spinner at a school fair has 8 equal sectors. One sector pays a prize of $20, two sectors pay $5 each, and the other five pay nothing. Let X be the prize won on one spin, in dollars. (a) Find E(X), and state the fee per spin that would make the game fair. (b) The stall charges $5 a spin and wants to expect a profit of $1 a spin. Keeping the two $5 sectors, what should the top prize be?

5/8$02/8$51/8$200: 5/8, 5: 2/8, 20: 1/85/8 + 2/8 + 1/8 = 1
The 8 sectors are equally likely: 5 pay nothing, 2 pay $5 and 1 pays $20.
Write the probability distribution of X. The sectors are equally likely, so P(X = 0) = 58, P(X = 5) = 28 and P(X = 20) = 18. Check: 58 + 28 + 18 = 1.
step 1 of 5

X is a random variable: it takes each of its values with a known probability. Its expectation E(X) is the average prize over a very large number of spins, found by multiplying every value by its probability and adding. A game is fair when the fee equals the expected prize, so that neither the player nor the stall expects to gain.

  1. Write the probability distribution of X. The sectors are equally likely, so P(X = 0) = 58, P(X = 5) = 28 and P(X = 20) = 18. Check: 58 + 28 + 18 = 1.
  2. Multiply each value by its probability and add: E(X) = 0 × 58 + 5 × 28 + 20 × 18 = 10 + 208 = 308 = 3.75.
  3. (a) E(X) = $3.75. The game is fair when the fee equals the expected prize, so a fair fee is $3.75 a spin.
  4. For a profit of $1 a spin at a fee of $5, the expected prize must be 5 − 1 = $4. Let the top prize be $x. Then E(X) = x + 2 × 58 = x + 108, so x + 108 = 4.
  5. Multiply both sides by 8: x + 10 = 32, so x = 22. (b) The top prize should be $22. Check: 22 + 108 = 4, and 5 − 4 = 1 dollar of profit a spin.

answer(a) E(X) = $3.75, so a fee of $3.75 is fair; (b) a top prize of $22

techniqueExpected Value · Random Variables · Probability Distributions

examsH2

Common pitfalls

  • Averaging the three prizes, 0 + 5 + 203 = 8.33. The three prizes are not equally likely: five of the eight sectors pay nothing, so each value must be weighted by its probability.
  • Treating the expected profit as a promise on every spin. One player wins $0, $5 or $22; the $1 is the average profit over many spins.
02

Deliveries to a Bakery Each Morning, Where One Probability Is Missing and the Spread Is Asked For

methodMake the Probabilities Add to 1, Find the Mean, Then Take the Mean of the Squares Minus the Square of the Mean

The number of deliveries X that a bakery receives in one morning has P(X = 0) = 0.1, P(X = 1) = 0.3, P(X = 2) = p and P(X = 3) = 0.2, and no other values are possible. (a) Find p and E(X). (b) Find Var(X) and the standard deviation of the number of deliveries.

0.100.31p20.230.1 + 0.3 + p + 0.2 = 1p = 1 − 0.6 = 0.4
The four probabilities add up to 1, so the missing bar is p = 0.4.
The probabilities add up to 1: 0.1 + 0.3 + p + 0.2 = 1, so p = 1 − 0.6 = 0.4.
step 1 of 5

The probabilities of all the values of a random variable add up to 1, which fixes the missing one. The mean E(X) weights each value by its probability. The variance measures the spread about the mean, and the quickest way to it is the mean of the squares minus the square of the mean: Var(X) = E(X2) − [E(X)]2.

  1. The probabilities add up to 1: 0.1 + 0.3 + p + 0.2 = 1, so p = 1 − 0.6 = 0.4.
  2. (a) p = 0.4, and E(X) = 0 × 0.1 + 1 × 0.3 + 2 × 0.4 + 3 × 0.2 = 0 + 0.3 + 0.8 + 0.6 = 1.7 deliveries.
  3. Find the mean of the squares, each square weighted by its own probability: E(X2) = 02 × 0.1 + 12 × 0.3 + 22 × 0.4 + 32 × 0.2 = 0 + 0.3 + 1.6 + 1.8 = 3.7.
  4. Subtract the square of the mean: Var(X) = 3.7 − 1.72 = 3.7 − 2.89 = 0.81.
  5. (b) Var(X) = 0.81, and the standard deviation is √0.81 = 0.9 deliveries. Check from the distances to the mean: 2.89 × 0.1 + 0.49 × 0.3 + 0.09 × 0.4 + 1.69 × 0.2 = 0.289 + 0.147 + 0.036 + 0.338 = 0.81.

answer(a) p = 0.4 and E(X) = 1.7; (b) Var(X) = 0.81, standard deviation 0.9 deliveries

techniqueVariance · Probability Distributions · Expected Value

examsH2

Common pitfalls

  • Writing Var(X) = E(X2) − E(X) = 3.7 − 1.7 = 2. The mean must be squared before it is subtracted: 1.72 = 2.89.
  • Squaring the probabilities instead of the values in E(X2). The value 2 becomes 4, and it is still weighted by its own probability, 0.4 and not 0.16.
03

A Plumber's Bill Made of a Call-Out Fee and an Hourly Rate, Compared with a Second Firm

methodWrite the Bill as aX + b; the Fixed Fee Moves the Mean but Not the Spread, and the Rate Scales Both

The time X hours that a plumber spends on a repair has E(X) = 1.5 and Var(X) = 0.25. Firm A charges a call-out fee of $40 plus $60 an hour, so its bill is C = 60X + 40 dollars. (a) Find the mean and the standard deviation of a bill from firm A. (b) Firm B charges $80 plus $45 an hour for the same repairs. Find the mean and the standard deviation of its bill, and say which firm is cheaper on average and which has the less variable bill.

hours X01.53firm A, $40130220E(C) = 60 × 1.5 + 40 = 130
Each hour of work maps to a bill. The mean time, 1.5 hours, maps to the mean bill: E(C) = 60 × 1.5 + 40 = 130.
For firm A, E(C) = 60E(X) + 40 = 60 × 1.5 + 40 = 90 + 40 = 130.
step 1 of 5

A bill of the form aX + b is a linear change of the variable. Its mean follows the same rule as each bill: E(aX + b) = aE(X) + b. A fixed fee moves every bill by the same amount, so it changes none of the distances between bills, while the rate a stretches every distance: Var(aX + b) = a2 Var(X), and the standard deviation is multiplied by a.

  1. For firm A, E(C) = 60E(X) + 40 = 60 × 1.5 + 40 = 90 + 40 = 130.
  2. The fee of $40 does not change the spread, and the rate squares in the variance: Var(C) = 602 × Var(X) = 3600 × 0.25 = 900.
  3. (a) A bill from firm A has mean $130 and standard deviation √900 = $30. Check: the standard deviation of X is √0.25 = 0.5 hours, and 60 × 0.5 = 30.
  4. Firm B's bill is 45X + 80. Its mean is 45 × 1.5 + 80 = 67.5 + 80 = $147.50, and its standard deviation is 45 × 0.5 = $22.50.
  5. (b) Firm A is cheaper on average, $130 against $147.50. Firm B's bill is less variable, with a standard deviation of $22.50 against $30, because its hourly rate is lower; its larger call-out fee adds nothing to the spread.

answer(a) mean $130, standard deviation $30; (b) firm B: mean $147.50, standard deviation $22.50; firm A is cheaper on average and firm B is less variable

techniqueTransforming a Random Variable · Expected Value · Variance

examsH2

Common pitfalls

  • Adding the fee to the variance, Var(C) = 900 + 40. Adding $40 to every bill moves them all together, so the spread is unchanged.
  • Multiplying the variance by 60 rather than 602, which gives 15 and a standard deviation of about $3.87. It is the standard deviation that is multiplied by 60, so the variance is multiplied by 3600.
04

Seeds That Fail to Come Up in a Tray of Twelve, and a Box of Five Trays That All Keep a Promise

methodName the Trial, n and p, Add the Binomial Terms Up to the Count Allowed, Then Treat Each Tray as a Trial of Its Own

Each seed in a nursery's trays comes up with probability 0.9, independently of the others. A tray holds 12 seeds, and the nursery promises at least 10 seedlings in every tray. Let X be the number of seeds in a tray that fail. (a) Find the probability that a tray keeps the promise, P(X ≤ 2). (b) A customer buys a box of 5 trays, whose seeds come up independently of one another too. Find the probability that all five trays keep the promise.

0123456X ~ B(12, 0.1): X counts failuresat least 10 seedlings: X <= 2
X ∼ B(12, 0.1) counts the seeds that fail. The bars for 7 to 12 failures are too small to show. At least 10 seedlings is X ≤ 2, the first three bars.
There are 12 independent trials, each seed fails with the same probability 0.1, and X counts the failures. So X ∼ B(12, 0.1), and P(X = r) = 12r (0.1)r (0.9)12 − r.
step 1 of 5

Each seed is a Bernoulli trial: it fails or it comes up. With a fixed number of independent trials and the same probability of failure each time, the number of failures is binomial, and P(X = r) = nr pr (1 − p)n − r. At least 10 seedlings means at most 2 failures, so three terms are added. A tray then either keeps the promise or does not, which is a trial of its own.

  1. There are 12 independent trials, each seed fails with the same probability 0.1, and X counts the failures. So X ∼ B(12, 0.1), and P(X = r) = 12r (0.1)r (0.9)12 − r.
  2. No failures: P(X = 0) = 0.912 = 0.2824. One failure, which can be any of the 12 seeds: P(X = 1) = 12 × 0.1 × 0.911 = 0.3766.
  3. Two failures, which can be any of 122 = 66 pairs of seeds: P(X = 2) = 66 × 0.12 × 0.910 = 66 × 0.01 × 0.3487 = 0.2301.
  4. (a) The tray keeps the promise when at most 2 seeds fail: P(X ≤ 2) = 0.2824 + 0.3766 + 0.2301 = 0.8891.
  5. The five trays are independent, and each keeps the promise with probability 0.8891. (b) The probability that all five do is 0.88915 = 0.556, to 3 significant figures. This is much lower than for one tray, because each of the five trays can break the promise.

answer(a) 0.8891; (b) 0.556

techniqueThe Binomial Distribution · Bernoulli Trials

examsH2

Common pitfalls

  • Finding only P(X = 2). At least 10 seedlings allows 0, 1 or 2 failures, so all three terms are added.
  • Leaving out 122. The product 0.12 × 0.910 is the chance that one particular pair of seeds fails and the rest come up, and there are 66 such pairs.
05

An Airline That Sells More Tickets Than Seats, Planned from the Binomial Mean and Standard Deviation

methodModel the Passengers Who Turn Up as Binomial, Use np and np(1 - p), Then Solve the Planning Rule as a Quadratic in the Square Root of n

An airline finds that 90% of passengers who book a flight turn up, independently of one another. (a) For a flight with 100 tickets sold, find the mean and the standard deviation of the number of passengers who turn up. (b) For an aircraft with 372 seats, the airline sells as many tickets as it can while keeping the mean plus two standard deviations of the number who turn up no more than 372. How many tickets does it sell?

80859095100100 tickets: number who turn upX ~ B(100, 0.9)
For 100 tickets the number who turn up is X ∼ B(100, 0.9), drawn here from 80 to 100.
For 100 tickets the number who turn up is X ∼ B(100, 0.9), since there are 100 independent trials, each with the same probability 0.9.
step 1 of 5

Each ticket is a trial: the passenger turns up or does not, independently, with the same probability 0.9. The number who turn up is therefore binomial, with mean np and variance np(1 − p). For n tickets both depend on n, so the planning rule becomes an equation in n, a quadratic in √n.

  1. For 100 tickets the number who turn up is X ∼ B(100, 0.9), since there are 100 independent trials, each with the same probability 0.9.
  2. (a) E(X) = np = 100 × 0.9 = 90 passengers, and Var(X) = np(1 − p) = 100 × 0.9 × 0.1 = 9, so the standard deviation is √9 = 3 passengers.
  3. For n tickets the mean is 0.9n and the standard deviation is √0.09n = 0.3√n. The rule asks for 0.9n + 2 × 0.3√n ≤ 372, that is 0.9n + 0.6√n ≤ 372.
  4. Let u = √n, and solve 0.9u2 + 0.6u − 372 = 0: u = −0.6 + √0.36 + 1339.21.8 = −0.6 + 36.61.8 = 20. The other root is negative and is rejected, because √n cannot be negative.
  5. So n = 202 = 400. (b) The airline sells 400 tickets. Check: the mean is 360 and the standard deviation √36 = 6, and 360 + 2 × 6 = 372. With 401 tickets the mean plus two standard deviations is about 372.9, which is over the seats.

answer(a) mean 90 passengers, standard deviation 3 passengers; (b) 400 tickets

techniqueBinomial Mean and Variance · Modeling with a Binomial

examsH2

Common pitfalls

  • Adding two variances instead of two standard deviations: 360 + 2 × 36 = 432. The rule is written in standard deviations, and 2 × 6 = 12.
  • Rounding up when the root is not a whole number. Every ticket beyond the largest n that satisfies the rule breaks it, so the answer is rounded down.
06

A Guessed Multiple-Choice Quiz That Is Binomial, and Question Cards Drawn from a Box That Are Not

methodCheck the Conditions for a Binomial Model Against Each Situation: a Fixed Number of Trials, Two Outcomes, Independent Trials and One Probability

A quiz has 5 questions, each with 4 options, and a student guesses every answer. (a) Explain why X, the number she gets right, can be modeled by B(5, 14), and find the probability that she gets at least 4 right. (b) In a later round the quiz-master draws 5 question cards at random, without replacement, from a box of 12 in which 3 are on sport. A friend models S, the number of sport questions, by B(5, 14). Explain why this model fails, and find P(S = 0) correctly and by the friend's model.

012345number right: B(5, 1/4)5 trials, right or wrong, independentp = 1/4 every time: B(5, 1/4)
The guesses meet all four conditions, so X ∼ B(5, 14).
There is a fixed number of trials, 5 questions. Each has two outcomes, right or wrong. The guesses are independent of one another, and each is right with the same probability 14. So X ∼ B(5, 14).
step 1 of 6

A binomial model needs four things: a fixed number of trials, two outcomes on each, trials that are independent, and the same probability of success on every trial. The guesses meet all four. The cards do not: once a card is drawn it is not put back, so what is left in the box, and the chance on the next draw, depends on the draws before it.

  1. There is a fixed number of trials, 5 questions. Each has two outcomes, right or wrong. The guesses are independent of one another, and each is right with the same probability 14. So X ∼ B(5, 14).
  2. Exactly 4 right, with the one wrong answer on any of the 5 questions: P(X = 4) = 54 (14)4 × 34 = 151024. All 5 right: P(X = 5) = (14)5 = 11024.
  3. (a) P(X ≥ 4) = 151024 + 11024 = 161024 = 164, about 0.0156.
  4. The cards are drawn without replacement. After a sport card, only 2 of the 11 cards left are on sport; after a card on another subject, 3 of the 11 are. The chance on each draw depends on the draws before it, so the draws are not independent and the binomial model fails.
  5. Correctly, all 5 cards come from the 9 that are not on sport: P(S = 0) = 95125 = 126792 = 744 ≈ 0.159.
  6. (b) The friend's model gives P(S = 0) = (34)5 = 2431024 ≈ 0.237, about half as large again as the true 0.159. The model fails because the draws are not independent.

answer(a) 164; (b) the draws are not independent, since each card drawn changes what is left in the box; P(S = 0) = 744 ≈ 0.159, against 2431024 ≈ 0.237 by the model

techniqueModeling with a Binomial · Bernoulli Trials · The Binomial Distribution

examsH2

Common pitfalls

  • Saying the model fails because p is not 14. Taken on its own, any one card is on sport with probability 312 = 14; what fails is independence, since each draw changes what the next one can be.
  • Leaving out 54 in P(X = 4) and writing 31024. The one wrong answer can be on any of the 5 questions, so there are 5 ways to get exactly 4 right.
07

Heights of Adult Women Modeled as Normal, a Proportion Between Two Heights and the Number Above One

methodStandardize Each Height to z, Read Φ from the Table, Use Symmetry for a Negative z, and Multiply a Probability by the Group for an Expected Count

The heights of adult women in a town are modeled by H ∼ N(165, 62), in centimeters. (a) Find the proportion of women whose heights lie between 159 cm and 174 cm. (b) A clothing shop stocks a tall range for women over 177 cm. Out of 500 women, how many would you expect to be over 177 cm?

cmz159−116501741.5N(165, 36)z = (h − 165)/6159 gives −1 and 174 gives 1.5
Under each height is its z-score, the number of standard deviations from the mean: 159 cm is at z = −1 and 174 cm at z = 1.5.
Standardize each height with z = h − 1656: 159 cm gives z = 159 − 1656 = −1, and 174 cm gives z = 174 − 1656 = 1.5.
step 1 of 5

Any normal variable is turned into the standard normal Z ∼ N(0, 1) by z = h − μσ, which counts standard deviations from the mean. The table gives Φ(z) = P(Z < z) for positive z; the curve is symmetric about 0, so Φ(−z) = 1 − Φ(z). An expected count is the probability multiplied by the size of the group.

  1. Standardize each height with z = h − 1656: 159 cm gives z = 159 − 1656 = −1, and 174 cm gives z = 174 − 1656 = 1.5.
  2. Read the table: Φ(1.5) = 0.9332. By symmetry, Φ(−1) = 1 − Φ(1) = 1 − 0.8413 = 0.1587.
  3. (a) P(159 < H < 174) = Φ(1.5) − Φ(−1) = 0.9332 − 0.1587 = 0.7745, so about 77% of the women.
  4. For 177 cm, z = 177 − 1656 = 2, and P(H > 177) = 1 − Φ(2) = 1 − 0.9772 = 0.0228.
  5. (b) The expected number out of 500 is 500 × 0.0228 = 11.4, so about 11 women. Check: 177 cm is 2 standard deviations above the mean, and a little over 2% of a normal distribution lies that far above its mean.

answer(a) 0.7745; (b) 11.4, so about 11 women

techniqueReading Normal Probabilities · Standardizing · Reading the Z-Table

examsH2

Common pitfalls

  • Dividing by the variance: 174 − 16536 = 0.25. N(165, 62) gives the variance 36; the z-score divides by the standard deviation, 6.
  • Reading Φ(−1) as 0.8413. The table holds positive z only, and 0.8413 is the area to the left of +1; the area to the left of −1 is 1 − 0.8413 = 0.1587.
08

A Machine Filling 500 g Bags of Rice: the Share That Is Underweight and the Mean It Must Be Set To

methodFor a Probability, Standardize the Mass; for a Setting, Read z Backwards from the Table and Solve for the Mean

A machine fills bags of rice labeled 500 g. The mass of rice in a bag is normally distributed with standard deviation 4 g, and the mean can be set. (a) With the mean set at 505 g, find the proportion of bags that are underweight. (b) The law allows at most 2.5% of bags to be under 500 g. Find the lowest mean the machine can be set to.

gz500−1.255050N(505, 16)z = (500 − 505)/4 = −1.25
With the mean at 505 g, the label mass 500 g sits at z = 500 − 5054 = −1.25.
Let M ∼ N(505, 42). A bag is underweight when M < 500, and z = 500 − 5054 = −1.25.
step 1 of 5

A probability is found by standardizing a value and reading Φ. The inverse question gives the probability and asks for a value: read the table backwards to find z, then solve z = x − μσ for the unknown, here the mean.

  1. Let M ∼ N(505, 42). A bag is underweight when M < 500, and z = 500 − 5054 = −1.25.
  2. (a) P(M < 500) = Φ(−1.25) = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056, so about 10.6% of bags are underweight.
  3. Let the new mean be μ. The table gives Φ(1.96) = 0.975, so the lowest 2.5% of a normal distribution lies below z = −1.96. The mass 500 g must sit at that z: 500 − μ4 = −1.96.
  4. Multiply both sides by 4: 500 − μ = −7.84, so μ = 507.84.
  5. (b) The lowest mean is 507.84 g, about 507.8 g. Check: 500 − 507.844 = −1.96, and Φ(−1.96) = 1 − 0.975 = 0.025.

answer(a) 0.1056; (b) 507.84 g

techniqueInverse Normal · The Normal Distribution · Standardizing

examsH2

Common pitfalls

  • Using z = +1.96, which gives μ = 500 − 7.84 = 492.16 g. A mean below the label would make most bags underweight; 500 g lies below the mean, so its z is negative.
  • Using z = −1.645, the value for 5% in one tail. The law allows 2.5% below 500 g, all in the lower tail, and that needs z = −1.96.
09

Waiting Time at a Walk-In Clinic with a Triangular Density, a Long Wait and the Median Wait

methodMake the Total Area Under the Density Equal 1, Read Every Probability as an Area, and Find the Median Where the Area Splits in Half

The time W minutes that a patient waits at a walk-in clinic has probability density f(w) = k(10 − w) for 0 ≤ w ≤ 10, and f(w) = 0 otherwise. (a) Find k, and the probability that a patient waits more than 6 minutes. (b) Find the median waiting time.

f(w)w10k010area = 1/2 × 10 × 10k = 50k = 1k = 1/50 = 0.02
The density is a triangle of base 10 and height 10k, and its area must be 1: k = 0.02.
The graph of f is a triangle with base 10 and height f(0) = 10k. The total area is 1: 12 × 10 × 10k = 50k = 1, so k = 150 = 0.02.
step 1 of 5

For a continuous random variable the probability of a range of values is the area under the density over that range, and the total area is 1. Here the graph of f is a straight line falling to 0 at w = 10, so every area is a triangle. The median m is the value with half of the area on each side.

  1. The graph of f is a triangle with base 10 and height f(0) = 10k. The total area is 1: 12 × 10 × 10k = 50k = 1, so k = 150 = 0.02.
  2. A wait of more than 6 minutes is the small triangle from w = 6 to w = 10. Its height is f(6) = 0.02 × 4 = 0.08, so its area is 12 × 4 × 0.08 = 0.16.
  3. (a) k = 0.02 and P(W > 6) = 0.16.
  4. The area to the right of the median m is a triangle with base 10 − m and height 0.02(10 − m), and it must be 0.5: 12 × 0.02(10 − m)2 = 0.5, so (10 − m)2 = 50.
  5. Then 10 − m = √50 = 7.071, so m = 2.929; the negative root would give m = 17.07, outside 0 ≤ w ≤ 10, and is rejected. (b) The median wait is 2.93 minutes, below 5 because short waits are the most likely. Check: 0.01 × 7.0712 = 0.500.

answer(a) k = 0.02 and P(W > 6) = 0.16; (b) 2.93 minutes

techniqueContinuous Variables · Random Variables

examsH2

Common pitfalls

  • Setting the height f(0) = 1, which gives k = 0.1. The height of a density is not a probability; it is the area under the graph that must equal 1.
  • Taking the median as 5 minutes, the middle of the range. The density is highest for short waits, so half of the area is used up well before 5 minutes.
10

Filled Jam Jars on a Packing Line: the Total Mass of Jar and Jam, and Lids That Must Be Wider Than the Neck

methodAdd the Means; Add the Variances Whether the Variables Are Added or Subtracted; Then Standardize the New Normal Variable

At a jam factory the mass of an empty jar is J ∼ N(200, 32) and the mass of jam put into it is M ∼ N(450, 42), in grams, independently. (a) Find the probability that a filled jar has a total mass under 640 g. (b) The inside diameter of a lid is L ∼ N(71.25, 0.42) and the outside diameter of a jar's rim is R ∼ N(70.0, 0.32), in millimeters, independently. A lid fits only if it is wider than the rim. Find the probability that a lid chosen at random does not fit a jar chosen at random.

gz640−26500T = J + M: N(650, 25)mean: 200 + 450 = 650Var: 9 + 16 = 25, sd 5
The means add and, for independent masses, so do the variances: T ∼ N(650, 52).
The total mass is T = J + M, with E(T) = 200 + 450 = 650. For independent variables the variances add: Var(T) = 32 + 42 = 9 + 16 = 25, so T ∼ N(650, 52).
step 1 of 5

A sum or a difference of independent normal variables is normal. The means add for a sum and subtract for a difference, but the variances ADD in both cases, because each variable brings its own spread. The new variable is then standardized in the usual way.

  1. The total mass is T = J + M, with E(T) = 200 + 450 = 650. For independent variables the variances add: Var(T) = 32 + 42 = 9 + 16 = 25, so T ∼ N(650, 52).
  2. (a) z = 640 − 6505 = −2, so P(T < 640) = 1 − Φ(2) = 1 − 0.9772 = 0.0228.
  3. The lid does not fit when L − R < 0. Let D = L − R; its mean is E(D) = 71.25 − 70.0 = 1.25 mm.
  4. The variances add for a difference too: Var(D) = 0.42 + 0.32 = 0.16 + 0.09 = 0.25, so D ∼ N(1.25, 0.52).
  5. (b) z = 0 − 1.250.5 = −2.5, so P(D < 0) = 1 − Φ(2.5) = 1 − 0.9938 = 0.0062. About 6 lids in 1000 will not fit a jar picked at random.

answer(a) 0.0228; (b) 0.0062

techniqueCombining Normals · Reading Normal Probabilities

examsH2

Common pitfalls

  • Subtracting the variances for L − R: 0.16 − 0.09 = 0.07. A difference is as uncertain as both of its parts together, so the variances add.
  • Adding the standard deviations, 3 + 4 = 7 g. Only variances add; the standard deviation of the total is √25 = 5 g.
Mr. Chalk Read the guide