Mixed practice: normal combinations, sample means and the central limit theorem Problem set A One question at a time

X and Y are independent, \(X\sim N(34,\,11)\) and \(Y\sim N(21,\,6)\). \(X_1,X_2\) are independent observations of X and \(Y_1,Y_2,Y_3\) are independent observations of Y. State the distribution of each.

1. (a) \(W=5X-2Y+3\)

   • Mean: \(E(W)\) = 5 × 34 − 2 × 21 + 3 = 131
   • Variance: \(Var(W)\) = 5² × 11 + (−2)² × 6 = 299

2. (b) \(W=X_1-X_2\)

   • Mean: 34 − 34 = 0
   • Variance: 1² × 11 + (−1)² × 11 = 22

3. (c) \(W=Y_1+Y_2+Y_3\)

   • Mean: 3 × 21 = 63
   • Variance: 3 × 6 = 18 (three observations, not 3²)

The mass of an item is a normal variable. Samples of independent items are taken and \(\bar X\) is the sample mean.

1. (a) The mass has distribution \(N(75,\,18)\). State the distribution of \(\bar X\) for a sample of 6 items.

   • \(Var(\bar{X})\) = 18 ÷ 6 = 3

2. (b) A second item has mass with mean 2.4 and standard deviation 0.5. Find the mean and the variance of the average mass of a batch of 10 such items.

   • Variance of one item = 0.5² = 0.25
   • Variance of the average = 0.25 ÷ 10 = 0.025

A random variable X has mean 45 and standard deviation 9. The distribution of X is not known. Give the approximate distribution of each quantity, or say that it cannot be found.

1. (a) \(\bar X\) for a sample of 60 observations

   • n = 60 > 30, so the central limit theorem applies
   • Variance = 9² ÷ 60 = 81 ÷ 60 = 1.35

2. (b) \(\bar X\) for a sample of 12 observations

   • n = 12 is not more than 30 and X is not known to be normal

3. (c) \(\sum_{i=1}^{40} X_i\)

   • n = 40 > 30, so the central limit theorem applies
   • Mean = 40 × 45 = 1800; variance = 40 × 9² = 3240

The mass of a loaf from a bakery is normally distributed with mean 800 g and standard deviation 25 g. Find the probability that a randomly chosen loaf has mass less than 760 g.

• \(X\sim N(800,\,25^2)\)
• From the calculator: \(P(X<760)\) = 0.0548 (normal cdf with μ = 800, σ = √625 = 25)

A courier van collects parcels. The mass of a parcel, in kg, is normally distributed with mean 18.4 and variance 5.2. The mass of its packaging, in kg, is normally distributed with mean 1.3 and variance 0.16. The two masses are independent.

1. (a) State the distribution of the total mass T and find any parameters.

   • Mean = 18.4 + 1.3 = 19.7
   • Variance = 5.2 + 0.16 = 5.36

2. (b) Find the probability that the total mass is more than 22 kg.

   • From the calculator: \(P(T>22)\) = 0.160 (normal cdf with μ = 19.7, σ = √5.36 = 2.32…)

Maya's time to solve a puzzle, A minutes, is normally distributed with mean 6.4 and variance 1.21. Noor's time, B minutes, is normally distributed with mean 7.1 and variance 0.81. The times are independent.

1. (a) Find the mean and the standard deviation of D = A − B.

   • Mean = 6.4 − 7.1 = −0.7
   • Variance = 1.21 + 0.81 = 2.02
   • Standard deviation = √2.02 = 1.42…

2. (b) Find the probability that A is less than B.

   • A < B means D < 0
   • From the calculator: \(P(D<0)\) = 0.689 (normal cdf with μ = −0.7, σ = √2.02 = 1.42…)

3. (c) Find the probability that A is less than B by more than 2.

   • A is less than B by more than 2 means D < −2
   • From the calculator: \(P(D<-2)\) = 0.180 (normal cdf with μ = −0.7, σ = √2.02 = 1.42…)

The length of a rope, X metres, is normally distributed with mean 3.1 and variance 0.09. The length of a chain, Y metres, is normally distributed with mean 1.4 and variance 0.04. Ropes and chains are chosen independently.

1. (a) Find the probability that a randomly chosen rope is more than twice as long as a randomly chosen chain.

   • X > 2Y means X − 2Y > 0
   • Mean = 3.1 − 2 × 1.4 = 0.3
   • Variance = 1² × 0.09 + (−2)² × 0.04 = 0.25
   • From the calculator: \(P(X-2Y>0)\) = 0.726 (normal cdf with μ = 0.3, σ = √0.25 = 0.5)

2. (b) Find the probability that the total length of 2 randomly chosen ropes and 3 randomly chosen chains is more than 10.5 metres.

   • Total = 2 observations of X and 3 of Y
   • Mean = 2 × 3.1 + 3 × 1.4 = 10.4
   • Variance = 2 × 0.09 + 3 × 0.04 = 0.3
   • From the calculator: \(P(T>10.5)\) = 0.428 (normal cdf with μ = 10.4, σ = √0.3 = 0.548…)

A person's reaction time is normally distributed with mean 0.31 seconds and standard deviation 0.05 seconds. A random sample of 12 reaction times is taken. Find the probability that the mean time is less than 0.29 seconds.

• \(\bar X\sim N(0.31,\,0.05^2/12)\)
• From the calculator: \(P(\bar{X}<0.29)\) = 0.0829 (normal cdf with μ = 0.31, σ = 0.0144…)

The daily sales of a shop have mean \$85 and standard deviation \$22, with an unknown distribution. Find the probability that the mean daily sales over a random sample of 45 days are between \$80 and \$90.

• n = 45 > 30, so the central limit theorem applies
• \(\bar X\approx N(85,\,22^2/45)\)
• From the calculator: \(P(80<\bar{X}<90)\) = 0.873 (normal cdf, μ = 85, σ = 3.28…)

The mass of a ceramic tile has mean 0.42 kg and standard deviation 0.05 kg. A box is packed with 60 randomly chosen tiles. Use the central limit theorem to find the probability that the total mass of the tiles is less than 25 kg.

• n = 60 > 30, so the central limit theorem applies
• \(S\approx N(60\times 0.42,\,60\times 0.05^2)\) = \(N(25.2,\,0.15)\)
• From the calculator: \(P(S<25)\) = 0.303 (normal cdf with μ = 25.2, σ = √0.15 = 0.387…)

The random variable X has mean 62 and standard deviation 14. A sample of 40 independent observations is taken. Find \(P(|\bar X-60|>4)\).

• n = 40 > 30, so \(\bar X\approx N(62,\,14^2/40)\)
• \(|\bar X-60|>4\) means \(\bar X>64\) or \(\bar X<56\)
• From the calculator: \(P(\bar{X}>64)\) = 0.183 (normal cdf with μ = 62, σ = √4.9 = 2.21…)
• From the calculator: \(P(\bar{X}<56)\) = 0.00336 (normal cdf with μ = 62, σ = √4.9 = 2.21…)
• Add the tails: 0.1831 + 0.003359 = 0.186

The mass of a tablet is normally distributed with mean 250 mg and variance 4 mg². Tablets are checked in batches of 8. A batch is rejected if the average mass is less than 248.8 mg or more than 251.2 mg.

1. (a) Find the mean and the variance of the average of a random sample of this size.

   • Mean = 250
   • Variance = 4 ÷ 8 = 0.5

2. (b) Hence find the probability that a batch is rejected.

   • From the calculator: \(P(\bar{X}<248.8)\) = 0.0448 (normal cdf with μ = 250, σ = √0.5 = 0.707…)
   • From the calculator: \(P(\bar{X}>251.2)\) = 0.0448 (normal cdf with μ = 250, σ = √0.5 = 0.707…)
   • Rejected: add the two tails, 0.04484 + 0.04484 = 0.0897

X and Y are independent, \(X\sim N(5,\,4)\) and \(Y\sim N(30,\,9)\). \(X_1,X_2\) are independent observations of X and \(Y_1,Y_2\) are independent observations of Y. Find \(P(Y_1-Y_2>X_1+X_2)\).

• Write the inequality as \(Y_1-Y_2-X_1-X_2>0\) and call the left side W.
• Mean = 30 − 30 − 5 − 5 = −10
• Variance = 9 + 9 + 4 + 4 = 26
• From the calculator: \(P(W>0)\) = 0.0249 (normal cdf with μ = −10, σ = √26 = 5.10…)