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qaws [65]
3 years ago
12

Question 8 -

Mathematics
1 answer:
FinnZ [79.3K]3 years ago
7 0

Answer:

Her wage on the 8th day is $105 ⇒ answer d

Step-by-step explanation:

* <em>Lets explain how to solve the problem</em>

- The average of set of data = The sum of data/the number of data

- The sum of data = the average of set of data × the number of data

∵ Cindy worked for 15 consecutive days

∵ She is earning an average wage of $91 per day

∴ Her total earning = 91 × 15 = $1365

∵ Her average earning during the first 7 days is $87 per day

∴ Her total earning in the first 7 days = 87 × 7 = $609

∵ Her average earning during the last 7 days is $93 per day

∴ Her total earning in the last 7 days = 93 × 7 = $651

- Her total earning in the 14 days is the sum of the earning in the

 first 7 days and the earning during the last 7 days

∵ Her total earning in the 14 days = 609 + 651 = $1260

∵ Her total earning in the 15 days is $1365

- To find her wage on the 8th day subtract her wage of the 14 days

  from her wage of the 15 days

∴ Her wage on the 8th day = 1365 - 1260 = $105

* Her wage on the 8th day is $105

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A selective college would like to have an entering class of 950 students. Because not all students who are offered admission acc
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Answer:

a) The mean is 900 and the standard deviation is 15.

b) 100% probability that at least 800 students accept.

c) 0.05% probability that more than 950 will accept.

d) 94.84% probability that more than 950 will accept

Step-by-step explanation:

We use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using 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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

(a) What are the mean and the standard deviation of the number X of students who accept?

n = 1200, p = 0.75. So

E(X) = np = 1200*0.75 = 900

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1200*0.75*0.25} = 15

The mean is 900 and the standard deviation is 15.

(b) Use the Normal approximation to find the probability that at least 800 students accept.

Using continuity corrections, this is P(X \geq 800 - 0.5) = P(X \geq 799.5), which is 1 subtracted by the pvalue of Z when X = 799.5. So

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

Z = \frac{799.5 - 900}{15}

Z = -6.7

Z = -6.7 has a pvalue of 0.

1 - 0 = 1

100% probability that at least 800 students accept.

(c) The college does not want more than 950 students. What is the probability that more than 950 will accept?

Using continuity corrections, this is P(X \geq 950 - 0.5) = P(X \geq 949.5), which is 1 subtracted by the pvalue of Z when X = 949.5. So

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

Z = \frac{949.5 - 900}{15}

Z = 3.3

Z = 3.3 has a pvalue of 0.9995

1 - 0.9995 = 0.0005

0.05% probability that more than 950 will accept.

(d) If the college decides to increase the number of admission offers to 1300, what is the probability that more than 950 will accept?

Now n = 1300. So

E(X) = np = 1300*0.75 = 975

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1200*0.75*0.25} = 15.6

Same logic as c.

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

Z = \frac{949.5 - 975}{15.6}

Z = -1.63

Z = -1.63 has a pvalue of 0.0516

1 - 0.0516 = 0.9484

94.84% probability that more than 950 will accept

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Answer: 140$

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eimsori [14]
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Answer:

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Step-by-step explanation:

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