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allsm [11]
2 years ago
15

Time left 0:43:35

Physics
1 answer:
ss7ja [257]2 years ago
7 0

Answer:

Bro where is image Hope you understand me

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Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl
ollegr [7]
<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

4 0
2 years ago
The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.320 with the floor. If
mart [117]

Answer:  29.50 m

Explanation: In order to calculate the higher accelation to stop a train  without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:

f=μ*N the friction force is equal to coefficient of static friction  multiply the normal force (m*g).

f=m.a=μ*N= m*a= μ*m*g= m*a

then

a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2

With this value we can determine the short distance to stop the train

as follows:

x= vo*t- (a/2)* t^2

Vf=0= vo-a*t then t=vo/a

Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m

5 0
2 years ago
A(n) 30 kg boy rides a roller coaster. The acceleration of gravity is 9.8 m/s 2 . With what force does he press against the seat
forsale [732]

Answer:

Force, F = 187.42 N

Explanation:

It is given that,

Mass of boy, m = 30 kg

Acceleration due to gravity, a=9.8\ m/s^2

Radius of curvature of the roller coaster, r = 15 m

Speed of the car, v = 7.3 m/s

The force acting on the boy are force of gravity and the centripetal force. The net force acting on him is as follows :

F=mg-\dfrac{mv^2}{r}

F=m(g-\dfrac{v^2}{r})

F=30\times (9.8-\dfrac{(7.3)^2}{15})

F = 187.42 N

So, he press against the seat with a force is 187.42 N. Hence, this is the required solution.

5 0
3 years ago
The output voltage of a power supply is normally distributed with mean 5 V and standard deviation 0.02 V. If the lower and upper
-BARSIC- [3]

To solve this problem we will apply the normal distribution, with which we will obtain the probability that the given event will occur. Concepts such as the mean and standard deviation will be present throughout the solution of the problem. Increasing or decreasing the average would change the location or center point of the curve. The change in the standard deviation would lead to the change in the dispersion of the data. As the standard deviation increases, the curve would become flatter.

Let X be the output voltage of power supply

X∼N (5,0.02^2)

A

The lower and upper specifications for voltage are 4.95 V and 5.05 V, respectively

P(4.95

P(4.95

P(4.95

P(4.95

P(4.95

Hence probability that a power supply selected at random will conform to the specifications on voltage is 0.9876

8 0
3 years ago
You have a series circuit powered by a 9V battery. If you double the amount of
ss7ja [257]
It gets 2 times weaker
5 0
2 years ago
Read 2 more answers
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