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Greeley [361]
3 years ago
8

Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl

e of pharmacy assistants.
Lower Bound Upper Bound
6.51 8.50
50 8.51
10.50 18
10.51 12.50
42 12.51
14.50 20
14.51 16.50
Physics
1 answer:
ollegr [7]3 years ago
4 0
<h2>Answer:</h2>

(a) standard deviation = σ = 4.9996

(b) variance = σ² = 24.996

<h2>Explanation:</h2><h2 />

<em>Given frequency table (find attached as Table 1);</em>

<u></u>

(a) To find the sample standard deviation and sample variance, follow these steps;

<em>i. Calculate the mid-point c for each group by using the mid-point formula;</em>

c = (lower bound + upper bound) / 2

=> c = (6.51 + 8.50) / 2 = 7.505

=> c = (8.51 + 10.50) / 2 = 9.505

=> c = (10.51 + 12.50) / 2 = 11.505

=> c = (12.51 + 14.50) / 2 = 13.505

=> c = (14.51 + 16.50) / 2 = 15.505

<em>So the new table becomes (find attached as Table 2);</em>

<em>ii. Calculate the total number of samples (n) which is the sum of all the frequencies.</em>

n = 50+18+42+20+46

n = 176

<em>iii. Calculate the mean (M)</em>

This is done by first multiplying the midpoints by the corresponding frequencies and then dividing the result by the total number of samples (n).

M = [(7.505 x 50) + (9.505 x 18) + (11.505 x 42) + (13.505 x 20) + (15.505 x 46)] / 176

M = [375.25 + 171.09 + 483.21 + 270.1 + 713.23] / 176

M = [2012.88] / 176

M = 11.44

<em>iv. Find the variance (σ²);</em>

The variance is calculated using the following formula

σ² = [Σ(f x c²) - (n x M²)] / (n - 1)                ------------(i)

Where;

f = frequency of each boundary data point

<em>=>  Let's first calculate </em>Σ(f x c²).

This is done by finding the sum of the product of the frequency (f) of each boundary point and the square of their corresponding mid-points(c)

Σ(f x c²) = [(50 x 7.505²) + (18 x 9.505²) + (42 x 11.505²) + (20 x 13.505²) + (46 x 15.505²)]

Σ(f x c²) = [(2816.25125) + (1626.21045) + (5559.33105) + (3647.7005) + (11058.63115)]

Σ(f x c²) = 24708.1244

<em>=> Now calculate (n x M²)</em>

n x M² = 176 x 11.44²

n x M² = 23033.7536

<em>=> Now substitute these values into equation (i) to calculate the variance</em>

σ² = [Σ(f x c²) - (n x M²)] / (n - 1)

σ² = [24708.1244 - 23033.7536] / (176 - 1)

σ² = [4374.3708] / (175)

σ² = 24.996

Therefore, the variance is 24.996

<em>v. Find the standard deviation (σ)</em>

The standard deviation is the square root of the variance. i.e

σ = √σ²

σ = √24.996

σ = 4.9996

Therefore, the standard deviation is 4.9996

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Anastaziya [24]

Answer:

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

Explanation:

||u|| = 70

θ = 40°

\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62

\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99

||v|| = 85

θ = 335°

\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03

\vec{v}_y=||v||sin\theta \\\Rightarrow \vec{v}_y=85sin335=-35.92

Resultant

R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15

\theta=tan^{-1}\frac{R_y}{R_x}\\\Rightarrow \theta=tan^{-1}\frac{70sin40+85sin335}{70cos40+85cos335}\\\Rightarrow \theta=tan^{-1}0.069=3.97^{\circ}

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

4 0
3 years ago
1. Calculate the momentum of each car before the collision: SHOW YOUR WORK!
a_sh-v [17]

Answer:

Momentum of red car = 5kgm/s

Momentum of blue car = 0kgm/s

Explanation:

Momentum = mass × velocity

For the red car

Mass = 1kg

Velocity = 5m/s

Momentum of the red car = 1kg × 5m/s

Momentum of the red car = 5kgm/s

For the blue car.

Mass = 1kg

Velocity = 0m/s(shows that the blue car is stationery)

Momentum = 1kg ×0m/s

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What are the conditions required for a rigid body to be in translational equilibrium?
kolezko [41]

Answer:

Explanation:

The condition for translation equilibrium is that is that the net force acting on the body must be zero.

The sum all the external forces acting on the body in horizontal as well as vertical direction must be zero.

∑Fₓ=0  and ∑Fy=0

now if the above two condition are satisfied the rigid body is said to be in  translational equilibrium.

God bless... hope this help to clear your doubt.

5 0
3 years ago
A gray kangaroo can bound across level ground with each jump carrying it 9.6 m from the takeoff point. Typically the kangaroo le
xz_007 [3.2K]

Answer:

(A) 11 m/s

(B) 1.3 m

Explanation:

Horizontal range, R = 9.6 m

Angle of projection, theta = 28 degree

(A)

Use the formula of horizontal range

R = u^2 Sin 2 theta / g

u^2 = R g / Sin 2 theta

u^2 = 9.6 × 9.8 / Sin ( 2 × 28)

u = 10.65 m/s

u = 11 m/s

(B)

Use the formula for maximum height

H = u^2 Sin ^2 theta / 2g

H =

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H = 1.275 m

H = 1 .3 m

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3 years ago
A spring is mounted horizontally, with its left end fixed. A spring balance attached to the free end and pulled toward the right
malfutka [58]

Answer:

200 N/m

20 rad/s

0.31415 seconds

3.18309 Hz

Explanation:

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x = Displacement of spring

F = Force on spring = 6 N

From Hooke's law we have relation

F=kx\\\Rightarrow k=\frac{F}{x}\\\Rightarrow k=\frac{6}{0.03}\\\Rightarrow k=200\ N/m

The spring constant is 200 N/m

Angular frequency is given by

\omega=\sqrt{\frac{k}{m}}\\\Rightarrow \omega=\sqrt{\frac{200}{0.5}}\\\Rightarrow \omega=20\ rad/s

The angular frequency is 20 rad/s

Frequency is given by

f=\frac{\omega}{2\pi}\\\Rightarrow f=\frac{20}{2\pi}\\\Rightarrow f=3.18309\ Hz

The frequency is 3.18309 Hz

Time period is given by

T=\frac{1}{f}\\\Rightarrow T=\frac{1}{3.18309}\\\Rightarrow T=0.31415\ s

The time period is 0.31415 seconds

6 0
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