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bagirrra123 [75]
2 years ago
5

At the beginning of the period, the Cutting Department budgeted direct labor of $46,300 and supervisor salaries of $37,200 for 4

,630 hours of production. The department actually completed 5,000 hours of production.
Mathematics
1 answer:
Korvikt [17]2 years ago
4 0

Answer:

$87200

Step-by-step explanation:

Here is the complete question:

At the beginning of the period, the Cutting Department budgeted direct labor of $46,300 and supervisor salaries of $37,200 for 4,630 hours of production. The department actually completed 5,000 hours of production.

Determine the budget of the department assuming that it uses flexible budgeting?

Given: Budget for direct labour= $46300

           Supervisor salaries= $37200

           Expected production hours= 4630 hours

           Completed production hours= 5000 hours

Now, we know that company budget include both fixed and variable cost.

∴ Direct labour cost is a variable cost and Supervisor salaries are fixed cost.

Using flexible budgeting for determining the budget of department, we will pro rate the direct labour cost on the basis of production hours.

Direct labour= Budget\times \frac{completed\ production\ hours}{expected\ production\ hours}

Direct labour= 46300\times \frac{5000}{4630}

∴ Direct labour= $50000

we know the department budget = Fixed cost+variable cost

∴ Department budget= \$ 37200+\$ 50000 = \$ 87200

∴ The department budget is $87200.

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No. 1.5 is a whole number while 1.5% is a percent
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2 years ago
The random variable X~(30,2^2) <br> Find p(X&lt;33) <br> Find p(X&gt;26)
Ostrovityanka [42]

Answer:

i) P(X<33)  = 0.9232

ii) P(X>26) = 0.001

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the mean of the Population = 30

Given that the standard deviation of the Population = 4

Let 'X' be the Normal distribution

<u>Step(ii):-</u>

i)

Given that the random variable  X = 33

               Z = \frac{x-mean}{S.D}

               Z = \frac{33-30}{2} = 1.5 >0

P(X<33) = P( Z<1.5)

              = 1- P(Z>1.5)

             = 1 - ( 0.5 - A(1.5))

             = 0.5 + 0.4232

  P(X<33)  = 0.9232

<u>Step(iii) :-</u>

Given that the random variable  X = 26

               Z = \frac{x-mean}{S.D}

               Z = \frac{33-26}{2} = 3.5 >0

P(X>26)  = P( Z>3.5)

              = 0.5 - A(3.5)

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2 years ago
A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat
zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

Due to the Central Limit Theorem, Z is:

Z = \frac{X - \mu}{s}

X = 205

Z = \frac{X - \mu}{s}

Z = \frac{205 - 200}{5}

Z = 1

Z = 1 has a pvalue of 0.8413.

X = 195

Z = \frac{X - \mu}{s}

Z = \frac{195 - 200}{5}

Z = -1

Z = -1 has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

X = 210

Z = \frac{X - \mu}{s}

Z = \frac{210 - 200}{5}

Z = 2

Z = 2 has a pvalue of 0.9772.

X = 195

Z = \frac{X - \mu}{s}

Z = \frac{190 - 200}{5}

Z = -2

Z = -2 has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

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3 years ago
Work out the value of 4 lots of 10​
nasty-shy [4]

Answer:

40

Step-by-step explanation:

4 lots of 10 is 10+10+10+10 or 4 x 10

3 0
2 years ago
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