Given:
Ten People
5 people = 30 years old
3 people = 32 years old
1 person = 31 years old
1 person = 65 years old

We need to arrange the ages in order from least to greatest.

30, 30, 30, 30, 30, 31, 32, 32, 32, 65
65 is the outlier.

A.) mean age without the outlier. count of the numbers : 9
5 x 30 = 150
1 x 31 = 31
3 x 32 =<u> 96</u>
total age 277
÷ # of pax<u> 9</u>
mean 30.77 round to the nearest tenth 30.80

b.) Standard deviation:
We need to calculate variance first:
5 pax : 30 - 30.80 = - 0.80 ⇒ -0.80² = 0.64 * 5pax = 3.20
1 pax : 31 - 30.80 = 0.20 ⇒ 0.20² = 0.04 * 1pax = 0.04
3 pax : 32 - 30.80 = 1.20 ⇒ 1.20² = 1.44 * 3 pax = 4.32

Variance : (3.20 + 0.04 + 4.32) / 9 pax = 7.56 / 9 = 0.84

Standard deviation is simply the square root of the variance.

Standard deviation = √0.84 = 0.916

0.916 round to nearest hundredths is 0.92
0.916 round to nearest tenths is 0.90

8 0

3 years ago

If a particle conned to move in one dimension has wavefunction ψ(x), then the probability of nding it between any two points x1

Andru [333]

Answer:

$P(0,a/3)=\frac{1}{3}-\frac{\sqrt{3}}{4\pi}$

Step-by-step explanation:

The probability of finding a particle between two points is given by:

$P(x_1,x_2)=\int_{x_1}^{x_2}|\Phi_n|^2dx$

Furthermore, the wave function of a nth state is given by:

$\Phi_n=\sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})$

By replacing we obtain

$P(0,\frac{x}{3})=\int_{0}^{\frac{a}{3}}\frac{2}{a}sin^2(\frac{n\pi x}{a})dx$

and by using the relation,

$sin^2\theta=\frac{1}{2}(1-cos(2\theta))$

we get for the ground state (n=1):

$P(0,a/3)=(\frac{2}{a})(\frac{1}{2})\int_{0}^{\frac{a}{3}}(1-cos(\frac{2\pi x}{a}))dx\\\\P(0,a/3)=(\frac{2}{a})(\frac{1}{2})[\frac{a}{3}-(\frac{a}{2\pi})sin(\frac{2\pi (a/3)}{a})]\\\\P(0,a/3)=(\frac{1}{a})[\frac{a}{3}-(\frac{a}{2\pi})(\frac{\sqrt{3}}{2})]=\frac{1}{3}-\frac{\sqrt{3}}{4\pi}$

hope this helps!!

7 0

3 years ago