Question

3. (a) The amount of orange juice in a ”Minute Made” carton is normally distributed with a mean of 312 millilitres (mL) and a standard deviation of 10 mL. Every ”Minute Made” carton is labelled with the serving size as 300 mL. i. What is the probability that a randomly selected carton has less than the labelled serving? [3 marks] ii. Determine the amount of orange juice in a carton for which only 2% of cartons fall below this amount. [3 marks] iii. Suppose that ”Minute-Made” company sells cartons of orange juice as a pack of 24 to grocery stores. What is the probability that the mean weight of a pack of 24 is in between 305mL and 320mL?

(b) In Singapore, only 40% of the adult citizens have a driving license. Suppose 1000 Singaporean adult citizens are randomly recruited. What is the probability that more than 420 of them possess a driving license? [5 marks]

(c) The San Antonio Spurs basketball team is one of 16 teams who made the NBA playoffs. During the playoffs, the Spurs needs to win over 3 other teams in order to get the championship match. In order to win over an opposing team, the Spurs need to win 4 out of 7 games in a match with the opposing team. Suppose there is a 0.8 probability for the Spurs to win a game. What is the probability that the Spurs can get to the championship match? [5 marks]

Answer 1

(3) Given = 312 and = 10

To find the probability, we need to find the z scores.

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(i) n = 1, P(X < 300

Z = (300 – 312) / 10 = -1.2

The required probability from the normal distribution tables is = 0.1151

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(ii) The value below X is less than 2%

To find P(X < x) = 0.02

From the standard normal distribution table, the z score at a p value of 0.02 Is -2.054

Therefore -2.054 = (X – 312) / 10

Solving for X, we get X = (-2.054 * 10) + 312 = 291.46

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(iii) n = 24

Between 305 and 320 = P(305 < X < 320) = P(X < 320) - P(X < 300)

For P(X < 320) ; z = (320 - 312) / [10/sqrt(24)] = 3.92. The p value at this score is = 1.0000

For P(X < 305) ; z = (305 - 312) / [10/sqrt(24)] = -3.43. The p value at this score is = 0.0003

Therefore the required probability is 1.0000 – 0.00003 = 0.9997

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(b) We use the Binomial Approximation to the normal as n is very large

Mean = n * p = 1000 * 0.4 = 400

Standard Deviation = Sqrt(n * p * q) = Sqrt(1000 * 0.4 * 0.6) = 15.492

P(X > 420). Using the continuity correction factor, we find the probability for P(X > 420+0.50 = P(X > 420.5)

For P (X > 420.5) = 1 - P (X < 420.5), as the normal tables give us the left tailed probability only.

For P( X < 420.5)

Z = (420.5 – 400) / 15.492 = 1.32

The probability for P(X < 420.5) from the normal distribution tables is = 0.9066

Therefore the required probability = 1 – 0.9066 = 0.0934

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Continuity Correction Factor Table

1) If P(X = n)       Use P(n – 0.5 < X < n+0.5)

2) If P(X > n)       Use P(X > n + 0.5)

3) If P(X <= n)    Use P(X < n + 0.5)

4) If P(X < n)       Use P(X < n - 0.5)

5) If P(X => n)    Use P(X > n - 0.5)

6) If P(a <= X < = b), Use P(X < b + 0.5) - P(X < a - 0.5)

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(c) Please note

Binomial Probability = ⁿC_x * (p)^x * (q)^n-x, where n = number of trials and x is the number of successes.

Here n = 16, p = 0.8, q = 1 – p = 0.2.

Also sum of probabilities from 0 till n = 1, i.eP(0) + P(1) + P(2) +.......+P(n) = 1

P(At least 3) = P(3) + P(4) + .... + P(16) = 1 - P(0) + P(1) + P(2)

P(X = 0) = ¹⁶C₀ * (0.8)⁰ * (0.2)^16-0 = 0.0000

P(X = 1) = ¹⁶C₀ * (0.8)¹ * (0.2)^16-1 = 0.0000

P(X = 2) = ¹⁶C₀ * (0.8)² * (0.2)^16-2 = 0.0000

Therefore P(At least 2) = 1 - (0 + 0 + 0) = 1.0000

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