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kaheart [24]
3 years ago
14

Arthur Industries entered into the following transactions during the month of June. Purchased supplies for $5,300 cash. Paid $4,

480 for salaries and wages for the month of June. Paid $560 in advance for July rent. Provided $13,400 in services on account. Paid $800 on accounts payable. Received $310 from customers as deposits for future services. Received a bill for $410 from the plumber who repaired a broken pipe in the restrooms, but will not pay the bill until July. Purchased equipment for cash of $740.
Business
1 answer:
VMariaS [17]3 years ago
8 0

Answer:

S/N   Account Titles and Explanation      Debit     Credit

A       Supplies                                             $5,300

              Cash                                                             $5,300

          (To record the purchase of supplies for cash)

B       Salaries and wages expense            $ 4,480

              Cash                                                             $4,480

         (To record the payment of wages and salaries)

C        Prepaid rent                                       $ 560

               Cash                                                             $560

          (To record the payment of prepaid rent for July)

D        Accounts receivable                           $13,400

                Service revenue                                         $13,400

         (To record the services provided on account)

E         Accounts payable                               $800

                   Cash                                                          $800

           (To record the payment made on Accounts payable)

F          Cash                                                     $310

                    Unearned revenue/Deferred revenue    $310

           (To record the unearned services revenue)

G          Repairs and maintenance expense  $410

                     Accounts payable                                     $410

             (To record the accounts payable for repairs expenses incurred)

H           Equipment                                          $740

                      Cash                                                           $740

             (To record the purchase of equipment for cash)

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Full Question

The united states department of agriculture (usda) found that the proportion of young adults ages 20–39 who regularly skip eating breakfast is 0.238. suppose that lance, a nutritionist, surveys the dietary habits of a random sample of size n=500 of young adults ages 20–39 in the united states.

Apply the central limit theorem for the binomial distribution to find the probability that the number of individuals, ?, in Lance's sample who regularly skip breakfast is greater than 122. You may find table of critical values helpful.

Express the result as a decimal precise to three places.

Answer:

The probability using the Normal Approximation that out of lance’s sample more than 122 people is 0.356

Explanation:

Given

n = Sample Size = 500

p = Probability = 0.238

Using the normal approximation to binomial distribution,

Let X = Event such that a person skips breakfast

If X ~ Binomial (n,p)

Using normal approximation

X ~ Normal (np,npq)

Where n = 500 and p = 0.238

So, X ~ Binomial (n,p) becomes

X ~ Binomial (500 , 0.238)

Using Normal Approximatiom

X ~ (119, 90.678)

The critical table is then constructed as follows;

Binomial --------- Normal

P(X = a) --------- P(a - 0.5 < X < a + 0.5)

P(X ≥ a) --------- P(X > a - 0.5)

P(X > a) --------- P(X > a + 0.5)

P(X ≤ a) --------- P(X < a + 0.5)

P(X < a) --------- P(X < a - 0.5)

Calculating the probability using the Normal Approximation that out of lance’s sample more than 122 people skips

This can be written as P(X > 122)

Looking at the critical table above.

P(X>a) ----- P(X>a + 0.5)

So,

P(X > 122) ---- P(X > 122 + 0.5)

P(X > 122) ---- P(X > 122.5)

Calculating Z score using

z = (x - u)/√σ²

X = 122.5

From X ~ (119, 90.678)

u =mean = 119

σ = standard deviation = √90.678

So,

Z = (122.5 - 119)/√90.678

z = 0.367550550865750

Z = 0.37 ---- Approximated

P(X > 122.5) = P(Z > 0.37)

P(X > 122.5) = 1 - P(Z<0.37) --- using z table

P(X > 122.5) = 1 - 0.6443

P(X > 122.5) = 0.3557

P(X > 122.5) = 0.356 -- Approximated

6 0
4 years ago
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