question 4\nbirths are approximately uniformly distributed between the 52 weeks of the year. they can be…

Explanation:

Step1: Recall mean formula for uniform distribution

For a uniform distribution $U(a,b)$, the mean $\mu=\frac{a + b}{2}$. Here $a = 1$ and $b=52$, so $\mu=\frac{1+52}{2}=26.5$.

Step2: Recall standard - deviation formula for uniform distribution

The standard deviation $\sigma=\sqrt{\frac{(b - a)^2}{12}}$. Substituting $a = 1$ and $b = 52$, we have $\sigma=\sqrt{\frac{(52 - 1)^2}{12}}=\sqrt{\frac{51^2}{12}}=\sqrt{\frac{2601}{12}}\approx14.9145$.

Step3: Probability at a single point for continuous distribution

For a continuous uniform distribution, the probability at a single point is $0$. So $P(x = 42)=0$.

Step4: Calculate $P(11\lt x\lt38)$

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 $f(x)=\frac{1}{52 - 1}=\frac{1}{51}$. Then $P(11\lt x\lt38)=\frac{38 - 11}{52 - 1}=\frac{27}{51}\approx0.5294$.

Step5: Calculate $P(x\gt31)$

$P(x\gt31)=\frac{52 - 31}{52 - 1}=\frac{21}{51}\approx0.4118$.

Step6: Calculate $P(x\gt18|x\lt35)$

Using the formula for conditional probability $P(A|B)=\frac{P(A\cap B)}{P(B)}$. Here $A={x\gt18}$ and $B = {x\lt35}$, so $A\cap B={18\lt x\lt35}$. Then $P(x\gt18|x\lt35)=\frac{35 - 18}{35 - 1}=\frac{17}{34}=0.5$.

Step7: Find the 28th percentile

Let the 28th percentile be $k$. We use the formula $P(x\lt k)=0.28$. Since $P(x\lt k)=\frac{k - 1}{52 - 1}$, then $\frac{k - 1}{51}=0.28$, and $k=1+0.28\times51=1 + 14.28 = 15.28$.

Step8: Find the minimum for the upper quartile

The upper quartile is the 75th percentile. Let it be $q$. Using the formula $P(x\lt q)=\frac{q - 1}{52 - 1}=0.75$. Then $q=1+0.75\times51=1 + 38.25 = 39.25$.

Answer:

a. $26.5000$ b. $14.9145$ c. $0.0000$ d. $0.5294$ e. $0.4118$ f. $0.5000$ g. $15.2800$ h. $39.2500$