Q: 30. Find the average of 6, 2.5, 3, 3, 2

The correct answer is: $3.3$ Explanation We find the average by summing the numbers and dividing by how many there are. Steps: Sum the numbers: $$6+2.5+3+3+2=16.5$$ Divide by the count (5): $$\frac{16.5}{5}=3.3$$ (Optional as a fraction) $$3.3=\frac{33}{10}$$ Therefore, the average is $3.3$.

Q: The mean is: The median is: The sample standard deviation is: The first quartile is: The third quartile is: What percent of the respondents have at least 3 siblings? 85% of all respondents have fewer than how many siblings?

I can’t compute those statistics without the data — please provide the list of sibling counts or a frequency table/histogram. Explanation / What I need To compute mean, median, sample standard deviation, Q1, Q3, and the requested percentiles I need either: the full list of each respondent’s number of siblings (e.g., 2,0,1,3,...) OR a frequency table (e.g., 0 siblings: 10 respondents; 1 sibling: 20 respondents; …) OR a clear image of the histogram with bin counts. Also tell me the sample size if it’s not obvious. How I will compute each item (so you know what I’ll do once you send the data) Mean: Formula: $\bar{x}=\dfrac{1}{n}\sum_{i=1}^n x_i$ Steps: sum all sibling counts, divide by $n$. Median: Sort the data. If $n$ is odd, median is the middle value. If $n$ is even, median is the average of the two middle values. Sample standard deviation: Formula: $s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n (x_i-\bar{x})^2}$ Steps: compute deviations from the mean, square them, sum, divide by $n-1$, take square root. First quartile (Q1) and third quartile (Q3): Method: sort the data. Q1 is the median of the lower half, Q3 is the median of the upper half. (Note: there are multiple conventions — I can use the Tukey method or the $ (n+1)p$ interpolation method; tell me if you need a specific convention.) Percent with at least 3 siblings: Compute count of respondents with $x\ge 3$, then percent$=100\times\frac{\#\{x\ge3\}}{n}$. “85% of all respondents have fewer than how many siblings?” (85th percentile) Find the 85th percentile (value $x$ such that 85% of observations are less than $x$). Methods: For discrete counts with a frequency table, find the smallest sibling count for which the cumulative percent is $\ge85\%$. For interpolation (continuous-style), use the position $p=(0.85)(n+1)$ and interpolate between sorted values. If you paste the raw data or a frequency table here, I will compute each statistic step-by-step and give the numeric answers.

Q: How many centimeters are equal to 1 meter?

The correct answer is: $100\text{ cm}$ Explanation The prefix centi- means one hundredth, so one centimeter is $0.01$ meters. Steps: By definition: $1\text{ cm} = 0.01\text{ m}$ Multiply both sides by $100$: $$100\text{ cm} = 1\text{ m}$$ Rearranging: $$1\text{ m} = 100\text{ cm}$$ Therefore, 1 meter equals 100 centimeters.

Q: 6 oz of soup = X ml

The correct answer is: 177.44 mL (for US fluid ounces) Explanation 6 ounces of soup typically means 6 US fluid ounces. Use the conversion $1\text{ fl oz} = 29.5735295625\text{ mL}$. Steps: Multiply: $$6\times29.5735295625=177.441177375\text{ mL}$$ Round to a sensible precision: 177.44 mL Note: If you meant 6 Imperial (UK) fluid ounces instead, use $1\text{ imp fl oz}=28.4130625\text{ mL}$, which gives $$6\times28.4130625=170.478375\text{ mL}\approx170.48\text{ mL}.$$

Q: How many months are equal to 45 days?

The correct answer is: $1.5\text{ months}$ Explanation This depends on the month length you assume. Commonly a month is approximated as 30 days, so: Steps: Using a 30-day month: $$\frac{45}{30}=1.5$$ Write that as months and days with a 30-day month: $1.5\text{ months}=1\text{ month }+0.5\times30\text{ days}=1\text{ month }+15\text{ days}$ If you use the average calendar month length (one year ≈ 365.25 days so an average month ≈ 30.4375 days), then:$$\frac{45}{30.4375}\approx1.478\text{ months}$$ Note: If you need an exact calendar answer, it depends on which specific months are involved (e.g., 45 days from Jan 1 is 1 month and 14 days later; from Feb 1 in a non‑leap year it's 1 month and 17 days).

Question

(a) Assuming each component's failure/success is independent of the others, what is the probability all three components fail, resulting in disaster for the flight? (b) What is the probability at least one of the components does not fail?

Accepted Answer

The probability all three fail is $p_1p_2p_3$.  The probability at least one does not fail is $1 - p_1p_2p_3$.

Explanation

(a) By independence, the probability that all three components fail is the product of their individual failure probabilities: $P(	ext{all fail})=p_1\cdot p_2\cdot p_3$.

(b) "At least one does not fail" is the complement of "all three fail", so

$P(	ext{at least one does not fail})=1-P(	ext{all fail})=1-p_1p_2p_3$.

Notes / variants

If each component has the same failure probability $p$, then $P(	ext{all fail})=p^3$ and $P(	ext{at least one does not fail})=1-p^3$.
You can also express part (b) in terms of success probabilities $s_i=1-p_i$ by expanding the complement if needed, but the compact complement form above is simplest.

(a) Assuming each component’s failure/success is independent of the others, what is the probability all three components fail, resulting in disaster for the flight? (b) What is the probability at least one of the components does not fail?

Explanation

Notes / variants