Normal Distribution - IB Mathematics Applications and Interpretation

1. Find the mean weight, $\mu$, of the apples.
   Answer: 152.4 g This requires solving a system of two equations. Let $W$ be the weight. 1. $P(W < 130) = 0.15$ $\Rightarrow$ z-score is $z_1 = \text{invNorm}(0.15) \approx -1.036$. So, $\frac{130 - \mu}{\sigma} = -1.036$ $\Rightarrow$ $130 - \mu = -1.036\sigma$ (Eq. 1). 2. $P(W > 180) = 0.10$ $\Rightarrow$ $P(W < 180) = 0.90$. z-score is $z_2 = \text{invNorm}(0.90) \approx 1.282$. So, $\frac{180 - \mu}{\sigma} = 1.282$ $\Rightarrow$ $180 - \mu = 1.282\sigma$ (Eq. 2). Subtract Eq. 1 from Eq. 2: $(180 - \mu) - (130 - \mu) = 1.282\sigma - (-1.036\sigma)$. $50 = 2.318\sigma$ $\Rightarrow$ $\sigma \approx 21.57$. Substitute $\sigma$ into Eq. 1: $130 - \mu = -1.036(21.57) \approx -22.35$. $\mu = 130 + 22.35 = 152.35$. The closest answer is 152.4 g.
2. What is the interquartile range (IQR) of the cholesterol levels, correct to one decimal place?
   Answer: 33.7 mg/dL The interquartile range (IQR) is the difference between the third quartile (Q3, 75th percentile) and the first quartile (Q1, 25th percentile). Using the inverse normal function on a GDC: Q1 = $\text{invNorm}(0.25, 190, 25) \approx 173.137$. Q3 = $\text{invNorm}(0.75, 190, 25) \approx 206.863$. IQR = Q3 - Q1 = $206.863 - 173.137 \approx 33.726$. To one decimal place, the IQR is 33.7 mg/dL.
3. Which statement correctly compares the student's relative performance on the two tests?
   Answer: She performed better in Chemistry because her z-score was higher. To compare relative performance, we calculate the z-score for each test. A higher z-score indicates a better performance relative to the group. Chemistry z-score: $z_{chem} = \frac{88 - 80}{5} = \frac{8}{5} = 1.6$. Physics z-score: $z_{phys} = \frac{84 - 75}{6} = \frac{9}{6} = 1.5$. Since $1.6 > 1.5$, her performance was relatively better in Chemistry.
4. A carton is considered 'underfilled' if it contains less than the stated volume. What is the probability that a randomly chosen carton is underfilled?
   Answer: 0.266 We need to find the probability $P(V < 1000)$ for a normal distribution with $\mu = 1005$ and $\sigma = 8$. Using a GDC: $P(V < 1000) = \text{normalcdf}(-\infty, 1000, 1005, 8)$. The z-score is $z = (1000-1005)/8 = -5/8 = -0.625$. The probability is $P(Z < -0.625) \approx 0.26598$. Rounded to three decimal places, the probability is 0.266.
5. What is the minimum score required to qualify for the next round?
   Answer: 66 To be in the top 30%, a participant must score at or above the 70th percentile. We need to find the score $X$ such that $P(\text{Score} < X) = 0.70$. Using a GDC's inverse normal distribution function with area = 0.70, $\mu = 60$, and $\sigma = 12$, we get: $X = \text{invNorm}(0.70, 60, 12) \approx 66.29$. The minimum integer score required is 66 (as scores are usually integers, and 66 is the first integer score fully in the top 30%). If we round, it is 66.
6. What is the standard deviation of the weights of the coffee bags, correct to two decimal places?
   Answer: 11.70 g Let $W$ be the weight. We are given $\mu = 500$ and $P(W > 515) = 0.10$. This implies $P(W < 515) = 0.90$. We find the z-score for a cumulative probability of 0.90: $z = \text{invNorm}(0.90, 0, 1) \approx 1.282$. Using the z-score formula $z = \frac{x - \mu}{\sigma}$, we get $1.282 = \frac{515 - 500}{\sigma}$. This simplifies to $1.282 = \frac{15}{\sigma}$. Solving for $\sigma$: $\sigma = \frac{15}{1.282} \approx 11.70$ g.
7. Based on this information, which of the following is the best estimate for the standard deviation of the plant heights?
   Answer: 5 cm The empirical rule for a normal distribution states that approximately 68% of the data falls within one standard deviation of the mean. The interval given, [35 cm, 45 cm], is symmetric around the mean of 40 cm. The distance from the mean to either end of the interval is $45 - 40 = 5$ cm or $40 - 35 = 5$ cm. Since this interval contains 68% of the data, this distance represents one standard deviation. Therefore, the best estimate for the standard deviation is 5 cm.
8. Find the value of $\mu$, the mean time spent in the supermarket, to the nearest minute.
   Answer: 75 We are given $P(T > 85) = 0.20$, which means $P(T < 85) = 0.80$. We need to find $\mu$ given $\sigma = 12$. First, find the z-score corresponding to a cumulative probability of 0.80: $z = \text{invNorm}(0.80, 0, 1) \approx 0.8416$. Now use the z-score formula: $z = \frac{x - \mu}{\sigma}$. $0.8416 = \frac{85 - \mu}{12}$. Solving for $\mu$: $12 \times 0.8416 = 85 - \mu \Rightarrow 10.099 = 85 - \mu \Rightarrow \mu = 85 - 10.099 \approx 74.9$. To the nearest minute, $\mu = 75$.
9. On what percentage of days is the rainfall between 2.0 mm and 5.0 mm?
   Answer: 65.6% We need to calculate $P(2.0 < R < 5.0)$ for a normal distribution with $\mu = 4.2$ and $\sigma = 1.5$. Using a GDC's normal cumulative distribution function: $\text{normalcdf}(2.0, 5.0, 4.2, 1.5) \approx 0.6563$. Converting this probability to a percentage gives $0.6563 \times 100\% \approx 65.6\%$.
10. If a police officer observes 500 cars, what is the expected number of cars that are travelling above the speed limit?
    Answer: 79 First, find the probability that a single car is travelling above the speed limit, $P(S > 100)$. Given $\mu = 90$ and $\sigma = 10$, we calculate $P(S > 100) = \text{normalcdf}(100, \infty, 90, 10)$. The z-score is $z = (100 - 90)/10 = 1$. $P(Z > 1) \approx 0.158655$. The expected number of speeding cars out of 500 is $n \times p = 500 \times 0.158655 \approx 79.3275$. The closest integer value is 79.
11. What is the probability that a randomly selected bolt is defective?
    Answer: 0.0124 A bolt is defective if $D < 9.90$ or $D > 10.10$. We can calculate these probabilities separately and add them. $P(D < 9.90) = \text{normalcdf}(-\infty, 9.90, 10.00, 0.04) \approx 0.00621$. $P(D > 10.10) = \text{normalcdf}(10.10, \infty, 10.00, 0.04) \approx 0.00621$. Due to symmetry, these probabilities are the same. The total probability of being defective is $0.00621 + 0.00621 = 0.01242$. Alternatively, we can find the probability of a non-defective bolt, $P(9.90 < D < 10.10) \approx 0.9876$, and subtract from 1: $1 - 0.9876 = 0.0124$.
12. What is the standard deviation of the weights of this species of fish?
    Answer: 60 g The z-score formula is $z = \frac{x - \mu}{\sigma}$. We are given $z = -2.5$, $x = 1050$, and $\mu = 1200$. We need to find $\sigma$. Substitute the given values into the formula: $-2.5 = \frac{1050 - 1200}{\sigma}$. $-2.5 = \frac{-150}{\sigma}$. Solving for $\sigma$: $\sigma = \frac{-150}{-2.5} = 60$. The standard deviation is 60 g.
13. Find the value of the mean $\mu$.
    Answer: 17.6 We are given $\sigma = 5$ and $P(X < 22) = 0.81$. We need to find $\mu$. First, find the z-score corresponding to a cumulative probability of 0.81: $z = \text{invNorm}(0.81, 0, 1) \approx 0.8779$. Now use the z-score formula: $z = \frac{x - \mu}{\sigma}$. $0.8779 = \frac{22 - \mu}{5}$. Solving for $\mu$: $5 \times 0.8779 = 22 - \mu \Rightarrow 4.3895 = 22 - \mu \Rightarrow \mu = 22 - 4.3895 \approx 17.61$. The closest value is 17.6.
14. What is the probability that the magnitude of the error is greater than 0.25°C?
    Answer: 0.211 The magnitude of the error being greater than 0.25°C means the error is either less than -0.25°C or greater than 0.25°C. Let $E$ be the error. We need to find $P(E < -0.25) + P(E > 0.25)$. $P(E > 0.25) = \text{normalcdf}(0.25, \infty, 0, 0.2) \approx 0.1056$. Due to symmetry, $P(E < -0.25)$ is also 0.1056. The total probability is $0.1056 + 0.1056 = 0.2112$. Rounded to three decimal places, the probability is 0.211.
15. Due to the symmetry of the normal distribution, what is the probability that the journey takes less than 35 minutes?
    Answer: 0.12 The mean is $\mu = 45$ minutes. The value 55 minutes is $55 - 45 = 10$ minutes above the mean. The value 35 minutes is $45 - 35 = 10$ minutes below the mean. Because the normal distribution is symmetric about the mean, the probability of being more than $k$ units above the mean is equal to the probability of being more than $k$ units below the mean. Therefore, $P(T > 55) = P(T < 35)$. Since $P(T > 55) = 0.12$, it follows that $P(T < 35) = 0.12$.
16. The airline charges extra for any suitcase weighing over 20.5 kg. What percentage of suitcases will be charged extra?
    Answer: 10.6% We need to calculate the probability $P(X > 20.5)$ for a normal distribution with $\mu = 18$ and $\sigma = 2$. Using a GDC: $P(X > 20.5) = \text{normalcdf}(20.5, \infty, 18, 2)$. The z-score is $z = (20.5 - 18) / 2 = 2.5 / 2 = 1.25$. $P(Z > 1.25) \approx 0.1056$. To express this as a percentage, we multiply by 100: $0.1056 \times 100\% = 10.6\%$.
17. To be eligible for a scholarship, a student must score in the top 8% of all test takers. What is the minimum score required to be eligible for the scholarship?
    Answer: 613 To be in the top 8%, a student's score must be at the 92nd percentile. We need to find the score $X$ such that $P(\text{Score} < X) = 0.92$. Using a calculator's inverse normal distribution function with area = 0.92, $\mu = 500$, and $\sigma = 80$, we get $X \approx 612.5$. The minimum score is 613 (rounded to the nearest integer).
18. What is the mean lifespan of this brand of smartphone battery, correct to one decimal place?
    Answer: 37.5 months Let $T$ be the lifespan. We are given $P(T < 30) = 0.03$ and $\sigma = 4$. We need to find $\mu$. First, find the z-score corresponding to a cumulative probability of 0.03: $z = \text{invNorm}(0.03, 0, 1) \approx -1.881$. Now use the z-score formula: $z = \frac{x - \mu}{\sigma}$. So, $-1.881 = \frac{30 - \mu}{4}$. Solving for $\mu$: $\mu = 30 - 4(-1.881) = 30 + 7.524 = 37.524$. To one decimal place, the mean lifespan is 37.5 months.
19. A sunflower is chosen at random. Given that its height is greater than 2.3 metres, what is the probability that its height is less than 2.8 metres?
    Answer: 0.788 We need to find the conditional probability $P(H < 2.8 | H > 2.3)$. The formula is $P(A|B) = \frac{P(A \cap B)}{P(B)}$. Here, $A \cap B$ is the event $2.3 < H < 2.8$. Using a GDC: $P(2.3 < H < 2.8) = \text{normalcdf}(2.3, 2.8, 2.5, 0.3) \approx 0.5889$. $P(H > 2.3) = \text{normalcdf}(2.3, 10^9, 2.5, 0.3) \approx 0.7475$. Therefore, the conditional probability is $\frac{0.5889}{0.7475} \approx 0.7878$, which rounds to 0.788.

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