NCEA Level 2 Mathematics and Statistics 2023 - Probability Methods (91267)

Flashcards covering tree diagrams, contingency tables, relative risk, conditional probability, and normal distribution problems from NZQA Level 2 Mathematics and Statistics 2023 (91267).

In the claw game scenario, what are the probabilities of dropping the toy, winning, and losing on Alex's first attempt?

The probabilities for the first game are:

• Drops toy: $15\% = 0.15$
• Wins toy: $\frac{1}{10} = 0.10$
• Loses toy: $1 - (0.15 + 0.10) = 0.75$

In the claw game scenario, what is the probability that the claw drops the toy on the first game and Alex wins on the second game?

$P(\text{Drop}_1 \cap \text{Win}_2) = 0.15 \times \frac{1}{9} = \frac{0.15}{9} = \frac{1}{60} \approx 0.0167$ (or $1.67\%$).

Alex visits the arcade with friends $70\%$ of the time and wins $10\%$ of those visits. When alone ($30\%$ of visits), she wins on about $\frac{1}{4}$ of visits. What is the overall probability that she wins the claw game on a visit?

$P(\text{Win}) = (0.70 \times 0.10) + (0.30 \times 0.25) = 0.070 + 0.075 = 0.145$ (or $14.5\%$).

From the 2021 NZ on Air survey of $1420$ participants ($251$ youth and $1169$ adults), $191$ youth watched streaming services. What is the probability that a randomly chosen participant is a youth who watched streaming services?

$P(\text{Youth} \cap \text{Streaming}) = \frac{191}{1420} \approx 0.1345$ (or approximately $13.45\%$).

Tao claims that because more adults ($610$) watch online videos than youth ($228$), adults are more likely to watch online videos than youth. Why is this claim incorrect?

Tao confuses frequency with probability. Likelihood must be evaluated by proportion:

• Youth: $\frac{228}{251} \approx 0.9084$ ($90.8\%$)
• Adults: $\frac{610}{1169} \approx 0.5218$ ($52.2\%$) Youth are significantly more likely to watch online videos ($90.8\% > 52.2\%$).

In the 2021 survey of $1420$ participants, $838$ watched online video, $724$ watched streaming services, and $795$ watched TV. What is the daily reach (overall percentage) for each media type?

• Online video: $\frac{838}{1420} \times 100\% \approx 59.0\%$
• Streaming services: $\frac{724}{1420} \times 100\% \approx 51.0\%$
• TV: $\frac{795}{1420} \times 100\% \approx 56.0\%$

In 2021, the proportion of youth watching streaming services daily was $\frac{191}{251} \approx 0.7610$ compared to $\frac{533}{1169} \approx 0.4560$ for adults. How does the relative risk support the claim that youth are about $70\%$ more likely to stream than adults?

$\text{Relative Risk} = \frac{0.7610}{0.4560} \approx 1.669$ This indicates youth are approximately $1.67$ times as likely, or about $66.9\%$ (approximately $70\%$) more likely to watch streaming services than adults.

In Table 2, $102$ of $249$ youth in 2016 watched streaming services daily, compared to $473$ of $706$ youth in 2022. How much higher was the percentage in 2022 than in 2016?

• 2016 percentage: $\frac{102}{249} \times 100\% \approx 41.0\%$
• 2022 percentage: $\frac{473}{706} \times 100\% \approx 67.0\%$
• Difference: $67.0\% - 41.0\% = 26.0\%$ percentage points higher.

Daily streaming time for youth is normally distributed with $\mu = 113\,\text{minutes}$ and $\sigma = 35\,\text{minutes}$. What is the probability that a randomly chosen youth watches under $92\,\text{minutes}$ a day?

Calculate the $Z$-score: $Z = \frac{92 - 113}{35} = \frac{-21}{35} = -0.60$$P(X < 92) = P(Z < -0.60) \approx 0.2743$ (or $27.43\%$).

Daily streaming time for youth follows $N(113, 35^2)$. What is the probability that a youth spends between $2$ and $3\,\text{hours}$ ($120$ to $180\,\text{minutes}$) streaming daily?

Standardize both endpoints: $Z_1 = \frac{120 - 113}{35} = 0.20$$Z_2 = \frac{180 - 113}{35} \approx 1.914$$P(120 < X < 180) = P(0.20 < Z < 1.914) \approx 0.9722 - 0.5793 = 0.3929$ (or approximately $0.393$).

Daily online video watching for youth is normally distributed with $\mu = 94\,\text{minutes}$ and $\sigma = 30\,\text{minutes}$. What is the interquartile range (IQR)?

For a normal distribution, the middle $50\%$ falls between $Z = -0.6745$ and $Z = 0.6745$: $IQR = 2 \times 0.6745 \times 30 \approx 1.349 \times 30 \approx 40.47\,\text{minutes}$ (or approximately $40.5\,\text{minutes}$).

State two reasons why a normal distribution may not be appropriate to model daily minutes spent watching online media by youth.

1. Negative values: A normal distribution extends infinitely to negative values, but viewing time cannot be negative ($\ge 0$).
2. Skewness: Viewing time is often right-skewed by a small number of heavy users watching for very long periods, rather than being symmetric.

Youth watched a daily average of $34\,\text{minutes}$ of TV overall, but $114\,\text{minutes}$ when excluding non-watchers. What effect does including non-watchers have on the mean and distribution?

• Mean: The large number of zero values significantly reduces the mean from $114\,\text{minutes}$ down to $34\,\text{minutes}$.
• Distribution: It introduces a large peak at $0$, making the distribution non-normal, heavily right-skewed, or zero-inflated.

Assuming daily TV viewing for youth watchers is normally distributed with $\mu = 114\,\text{minutes}$, and $12\%$ watch more than $150\,\text{minutes}$, what is the standard deviation $\sigma$?

From standard normal tables for an upper-tail area of $0.12$: $Z \approx 1.175$ Using $Z = \frac{X - \mu}{\sigma}$: $1.175 = \frac{150 - 114}{\sigma} = \frac{36}{\sigma} \implies \sigma = \frac{36}{1.175} \approx 30.64\,\text{minutes}$.