Let's solve both problems from Quiz #3 step-by-step. Here's the breakdown:

Problem 1

Find the mean, variance, and standard deviation of the probability distribution:

Step 1: Mean (Expected Value)

The formula for the mean ($\mu$) is:
$\mu = \sum (x \cdot P(x))$

$\mu = (22)(0.08) + (23)(0.19) + (24)(0.36) + (25)(0.25) + (26)(0.07) + (27)(0.05)$

$\mu = 1.76 + 4.37 + 8.64 + 6.25 + 1.82 + 1.35 = 24.19$

Mean = 24.19

Step 2: Variance

The formula for variance ($\sigma^2$) is:
$\sigma^2 = \sum [(x - \mu)^2 \cdot P(x)]$

We'll first compute $(x - \mu)^2$ for each $x$, and then multiply by $P(x)$.

$\sigma^2 = 0.3837 + 0.2691 + 0.0130 + 0.1640 + 0.2293 + 0.3948 = 1.4539$

Variance = 1.4539

Step 3: Standard Deviation

The formula for standard deviation ($\sigma$) is:
$\sigma = \sqrt{\sigma^2}$

$\sigma = \sqrt{1.4539} = 1.2066$

Standard Deviation = 1.21

Problem 2

The probabilities of 0, 1, 2, 3, 4, 5 students coming late to a class are given as follows:

Step 1: Mean (Expected Value)

$\mu = \sum (x \cdot P(x))$

$\mu = (0)(0.40) + (1)(0.22) + (2)(0.15) + (3)(0.10) + (4)(0.08) + (5)(0.05)$

$\mu = 0 + 0.22 + 0.30 + 0.30 + 0.32 + 0.25 = 1.39$

Mean = 1.39

Step 2: Variance

$\sigma^2 = \sum [(x - \mu)^2 \cdot P(x)]$