Question

the age of children in kindergarten on the first day of school is uniformly distributed between 4.73 and 5.72 years old. a first - time kindergarten child is selected at random. round answers to 4 decimal places if possible.
a. the mean of this distribution is
b. the standard deviation is
c. what is the probability that the child will be older than 5 years old?
d. the probability that the child will be between 5.13 and 5.33 years old is
e. if such a child is at the 79th percentile, how old is that child? years old.

Explanation:

Step1: Recall mean formula for uniform distribution

For a uniform distribution $U(a,b)$, the mean $\mu=\frac{a + b}{2}$, where $a = 4.73$ and $b=5.72$. So, $\mu=\frac{4.73+5.72}{2}=\frac{10.45}{2}=5.2250$.

Step2: Recall standard - deviation formula for uniform distribution

The standard deviation $\sigma=\sqrt{\frac{(b - a)^2}{12}}$. Substitute $a = 4.73$ and $b = 5.72$: $\sigma=\sqrt{\frac{(5.72 - 4.73)^2}{12}}=\sqrt{\frac{0.99^2}{12}}=\sqrt{\frac{0.9801}{12}}\approx0.2860$.

Step3: Calculate the probability of being older than 5

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, $f(x)=\frac{1}{5.72-4.73}=\frac{1}{0.99}$. The probability $P(X>5)=\frac{b - 5}{b - a}=\frac{5.72 - 5}{5.72-4.73}=\frac{0.72}{0.99}\approx0.7273$.

Step4: Calculate the probability of being between 5.13 and 5.33

$P(5.13

Step5: Calculate the 79th percentile

For a uniform distribution $U(a,b)$, the $p$ - th percentile $x_p=a+(b - a)p$. Here, $p = 0.79$, $a = 4.73$, and $b = 5.72$. So, $x_{0.79}=4.73+(5.72 - 4.73)\times0.79=4.73+0.99\times0.79=4.73 + 0.7821=5.5121$.

Answer:

a. 5.2250
b. 0.2860
c. 0.7273
d. 0.2020
e. 5.5121