Question

the age of children in kindergarten on the first day of school is uniformly distributed between 4.88 and 5.87 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 5.7 years old?
d. the probability that the child will be between 5.18 and 5.48 years old is
e. if such a child is at the 32nd percentile, how old is that child? years old.
hint:
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Explanation:

Step1: Recall mean formula for uniform distribution

For a uniform distribution $U(a,b)$, the mean $\mu=\frac{a + b}{2}$, where $a = 4.88$ and $b=5.87$. So, $\mu=\frac{4.88+5.87}{2}=\frac{10.75}{2}=5.3750$.

Step2: Recall standard - deviation formula for uniform distribution

The standard deviation $\sigma=\sqrt{\frac{(b - a)^2}{12}}$, with $a = 4.88$ and $b = 5.87$. Then $\sigma=\sqrt{\frac{(5.87 - 4.88)^2}{12}}=\sqrt{\frac{0.99^2}{12}}=\sqrt{\frac{0.9801}{12}}\approx\sqrt{0.081675}\approx0.2858$.

Step3: Calculate probability for $x>5.7$

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 $f(x)=\frac{1}{5.87 - 4.88}=\frac{1}{0.99}$. The probability $P(X>5.7)=\frac{5.87 - 5.7}{5.87 - 4.88}=\frac{0.17}{0.99}\approx0.1717$.

Step4: Calculate probability for $5.18

$P(5.18

Step5: Calculate the 32nd percentile

The formula for the $k$ - th percentile in a uniform distribution $U(a,b)$ is $x=a + k(b - a)$. For the 32nd percentile, $k = 0.32$. So $x=4.88+0.32\times(5.87 - 4.88)=4.88+0.32\times0.99=4.88 + 0.3168=5.1968$.

Answer:

a. $5.3750$
b. $0.2858$
c. $0.1717$
d. $0.3030$
e. $5.1968$