QCAA Mathematical Methods Continuous random variables and the normal distribution

This range falls between 2 and 4 standard deviations below the mean ($z = -4$ to $z = -2$). This represents the far left tail of the curve, which contains a negligible percentage of the population.

This range covers the area from 2 standard deviations below the mean up to the mean ($z = -2$ to $z = 0$). While substantial, it captures less area than an interval of the same width centered directly on the peak of the distribution.

This range falls between 1 and 3 standard deviations above the mean ($z = 1$ to $z = 3$). This represents the tapering right tail of the distribution, which contains significantly fewer teenagers than the central range.

This is the value of the constant $k$, not the probability $\Pr(X < 0.5)$. The constant $k = \frac{1}{2}$ is found by setting the total area under the PDF to 1.

This is only the probability $\Pr(0 \le X < 0.5)$. It incorrectly omits the probability $\Pr(-3 \le X < 0) = \frac{3}{4}$ which must be included since the lower bound of the domain is $-3$.

This is an incorrect calculation. A probability value cannot exceed 1, so $\frac{49}{48}$ is impossible for any valid probability density function.

Incorrect. This value represents the z-score ($z = 1.5$) from the standard normal distribution, not the standard deviation of $X$.

Incorrect. This would be the standard deviation if the z-score was 3, but the given z-score is 1.5.

Incorrect. This is the difference between the given value and the mean ($259 - 250 = 9$), which is the numerator of the z-score formula, not the standard deviation.

Incorrect. This value might result from an arithmetic error, such as incorrectly multiplying the difference from the mean by a factor instead of dividing by the z-score.

This is the total area under the probability density function curve, $\int_0^\infty f(x) dx = 1$, which verifies it is a valid PDF but does not give the expected value.

This is an incorrect evaluation of the expected value. The correct formula requires evaluating the integral $\int_0^\infty x f(x) dx$.

This value results from an incorrect integration or applying the wrong formula for the expected value of a continuous random variable.

This is the second moment, $E(X^2) = \int_0^\infty x^2 f(x) dx = 9$, rather than the expected value $E(X)$.

Incorrect. This value represents the mean of the distribution ($\mu = 2\sigma \approx 14.476$), not the standard deviation.

Incorrect. This is a distractor that likely results from an algebraic error or incorrectly setting up the relationship between the mean and standard deviation.

Incorrect. This value is obtained by incorrectly using a positive z-score ($1.2816$), which corresponds to $\Pr(X < 5.2) = 0.9$ rather than $\Pr(X > 5.2) = 0.9$.

Incorrect. This represents the mean ($\mu = 2\sigma$) calculated under the incorrect assumption that $\Pr(X < 5.2) = 0.9$.

Incorrect. This value does not make the total area under the probability density function equal to 1, likely resulting from an error in evaluating the trigonometric integrals.

Incorrect. This might result from incorrectly rationalizing the denominator or making an arithmetic error when solving $k(2 - \sqrt{2}) = 1$.

Incorrect. This is the value of the integral when $k=1$. Since the total area must be 1, $k$ must be the reciprocal of this value.

The area to the left of $z=1$ represents the proportion of men with pulse rates less than 78. Since the question asks for rates greater than 78, you need the area to the right.

A z-score of $z = -1$ corresponds to a pulse rate of $70 - 8 = 62$. This calculation would determine the proportion of men with pulse rates greater than 62, not 78.

This area represents the proportion of men with pulse rates between 62 and 78 (within one standard deviation of the mean), rather than the specific tail proportion greater than 78.

The z-score for this subject is $z = \frac{77 - 62}{22} \approx 0.68$. While this is a positive result, it is lower than the z-score for Subject B.

The z-score for this subject is $z = \frac{70 - 61}{15} = 0.60$. This indicates a lower relative standing compared to Subject B.

The z-score for this subject is $z = \frac{82 - 73}{20} = 0.45$. Even though the raw score (82) is the highest, the relative performance is actually the lowest of the four subjects.

This incorrect probability likely results from a calculation error or failing to properly account for both the upper and lower bounds of the inequality $\sin(2\theta) > 0.8$.

This option is incorrect and may stem from miscalculating the inverse sine function or incorrectly standardizing the normal distribution values.

This incorrect answer could result from forgetting to divide the angles by 2 when solving $2\theta$, leading to incorrect z-scores.

This is approximately $P(\theta < 26.57^\circ)$, which represents the probability of the ball falling short of 40 m due to the kick angle being too low, rather than traveling more than 40 m.

This value represents the probability for the interval $10 \leq t < 15$ only ($30/200 = 0.15$). You must also include the interval $15 \leq t < 20$.

This option likely confuses the frequency count (70) with the probability, or assumes a total frequency of 100. You must divide the specific frequency (70) by the total frequency (200).

This is the probability that the service took at least 10 minutes ($t \geq 10$), calculated as $\frac{30+40+100}{200} = 0.85$. The question excludes times of 20 minutes or more.