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7 days ago

Millner Corporation has provided the following data from its activity-based costing accounting system: Activity Cost Pool Total

Cost Total Activity Designing products $ 1,372,448 7,798 product design hours Setting up batches $ 33,300 740 batch set-ups Assembling products $ 126,160 6,640 assembly hours The activity rate for the "designing products" activity cost pool is closest to:

Business

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Net interest margin—often referred to as spread—is the difference between the rate banks pay on deposits and the rate they charg

Nady [3600]

Answer:

(a) $P(X\:>\:5.40)=0.9938$

(b) $P(X\:$

Explanation:

(a) Identify the z-value that represents 5.40 percent

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

$Z=\frac{5.40-4.15}{0.5}$

$Z=\frac{1.25}{0.5}=2.5$

Thus, a net interest margin of 5.40 percent stands at 2.5 standard deviations above the average.

From the standard normal distribution table, the area to the left of 2.5 is 0.9938. Hence, the likelihood of a randomly selected U.S. bank achieving a net interest margin greater than 5.40 percent is 1-0.9938=0.0062

(b) The z-value corresponding to 4.40 percent is $Z=\frac{4.40-4.15}{0.5}=0.5$ The net interest margin of 4.40 percent is situated at 0.5 standard deviation above the average.

According to the normal distribution table, the area to the left of 0.5 is 0.6915

Thus, the probability of a randomly chosen U.S. bank having a net interest margin below 4.40 percent equals 0.6915

Substituting 1.65 into the equation enables us to find X. $1.65=\frac{X-4.15}{0.5}$

$1.65\times 0.5=X-4.15$ $0.825=X-4.15$

$0.825+4.15=X$

$4.975=X$

For a bank that wishes for its net interest margin to fall below that of 95 percent of all U.S. banks, it should aim for a net interest margin of 4.975 percent.

6 0

1 month ago

The gross profit method of inventory valuation is not valid when a. there is substantial increase in the quantity of inventory d

stepan [3596]

The gross profit method of inventory valuation is invalid under the circumstance where the gross margin percentage experiences significant changes throughout the year.

6 0

1 month ago

The united states department of agriculture (usda) found that the proportion of young adults ages 20–39 who regularly skip eatin

Katen [3525]

Answer:

Probability = 0.3557

Explanation:

Data provided

Young adults age range = 20 to 39

Proportion skipping breakfast p = 0.238

Sample size n = 500

Objective

We aim to determine the probability that more than 122 of Lance's sampled individuals do not eat breakfast regularly.

Solution

Utilizing the Normal Approximation for the Binomial Distribution,

let's define our random variable as x:

x ~ Bin (n,p).............1

Using Normal Approximation gives us:

x ~ Normal Approx (np, npq).................2

This simplifies to:

x ~ (500, 0.238)

where we know q will be:

q = 1 - p

q = 1 - 0.238

q = 0.762.............3

Hence, we find:

x ~ Normal Approx (119, 90.678)

Now we want to compute P(X > 122).

First, we convert this to Z:

z = $\frac{x-\mu}{\sigma}$ ................4

Where the mean is $\mu$ = np

and standard deviation is $\sigma = \sqrt{npq}$

For P(X > 122, we evaluate:

$P(\frac{X-\mu}{\sigma}>\frac{122-119}{\sqrt{90.678}})$ ............5

This results in P(Z > 0.37).

Thus,

Probability = 1 - P(Z < 0.37)

Next, we'll refer to the z table:

Probability = 1-0.6443

Probability = 0.3557

7 0

2 months ago