Question

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.
a. the mean of this distribution is
b. the standard deviation is
c. the probability that wave will crash onto the beach exactly 2 seconds after the person arrives is p(x = 2) =
d. 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) =
e. 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) =
f. 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
g. 34% of the time a person will wait at least how long before the wave crashes in? seconds.
h. find the minimum for the upper quartile. seconds.
hint:
written hint +
helpful videos: probability +, conditional probability + conditional probability + percentiles +

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

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

Step7: Find the value for a given percentile

If $P(X\geq x)=0.34$, then $P(X

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