today, the waves are crashing onto the beach every 5.3 seconds. the times from when a person arrives at the…
today, the waves are crashing onto the beach every 5.3 seconds. the times from when a person arrives at the shoreline until a crashing wave is observed follows a uniform distribution from 0 to 5.3 seconds. round to 4 decimal places where possible.\na. the mean of this distribution is \nb. the standard deviation is \nc. the probability that wave will crash onto the beach exactly 2 seconds after the person arrives is p(x = 2) =\nd. the probability that the wave will crash onto the beach between 0.3 and 4.4 seconds after the person arrives is p(0.3 < x < 4.4) =\ne. the probability that it will take longer than 3.76 seconds for the wave to crash onto the beach after the person arrives is p(x > 3.76) =\nf. suppose that the person has already been standing at the shoreline for 0.8 seconds without a wave crashing in. find the probability that it will take between 2.1 and 2.9 seconds for the wave to crash onto the shoreline. 3276\ng. 34% of the time a person will wait at least how long before the wave crashes in? seconds.\nh. find the minimum for the upper quartile. seconds.\nhint:\nwritten hint +\nhelpful videos: probability +, conditional probability + conditional probability + percentiles +
Answer
Explanation:
Step1: Recall mean formula for uniform distribution
For a uniform distribution $U(a,b)$, the mean $\mu=\frac{a + b}{2}$. Here $a = 0$ and $b=5.3$, so $\mu=\frac{0 + 5.3}{2}=2.65$.
Step2: Recall standard - deviation formula for uniform distribution
The standard deviation $\sigma=\sqrt{\frac{(b - a)^2}{12}}$. Substituting $a = 0$ and $b = 5.3$, we get $\sigma=\sqrt{\frac{(5.3-0)^2}{12}}=\sqrt{\frac{28.09}{12}}\approx1.5270$.
Step3: Probability at a single point in continuous distribution
For a continuous uniform distribution, the probability at a single point is 0. So $P(X = 2)=0$.
Step4: Calculate probability for an interval
The probability density function of a uniform distribution $U(a,b)$ is $f(x)=\frac{1}{b - a}$ for $a\leq x\leq b$. Here $f(x)=\frac{1}{5.3-0}=\frac{1}{5.3}$. Then $P(0.3<X<4.4)=\frac{4.4 - 0.3}{5.3}=\frac{4.1}{5.3}\approx0.7736$.
Step5: Calculate right - tailed probability
$P(X>3.76)=\frac{5.3 - 3.76}{5.3}=\frac{1.54}{5.3}\approx0.2906$.
Step6: Calculate conditional probability
Since the uniform distribution is memory - less, the conditional distribution of $X$ given $X\geq0.8$ is also uniform on $[0.8,5.3]$. The new interval for the conditional probability is from $2.1$ to $2.9$. The length of the new sample space is $5.3 - 0.8 = 4.5$, and the length of the event space is $2.9 - 2.1=0.8$. So $P(2.1<X<2.9|X\geq0.8)=\frac{2.9 - 2.1}{5.3 - 0.8}=\frac{0.8}{4.5}\approx0.1778$.
Step7: Find the value for a given percentile
If $P(X\geq x)=0.34$, then $P(X<x)=1 - 0.34 = 0.66$. Using the formula for the cumulative - distribution function of a uniform distribution $F(x)=\frac{x - a}{b - a}$, we have $\frac{x-0}{5.3}=0.66$, so $x = 0.66\times5.3=3.498$.
Step8: Find the upper quartile
The upper quartile $Q_3$ corresponds to the 75th percentile. Using the formula $F(x)=\frac{x - a}{b - a}$, with $F(x)=0.75$, $a = 0$ and $b = 5.3$, we get $\frac{x-0}{5.3}=0.75$, so $x=0.75\times5.3 = 3.975$.
Answer:
a. 2.65 b. 1.5270 c. 0 d. 0.7736 e. 0.2906 f. 0.1778 g. 3.498 h. 3.975