The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probability of each of the following occurring from this population. Appendix A Statistical Tables a. A random sample of size 36 yielding a sample mean of 78 or more b. A random sample of size 150 yielding a sample mean of between 71 and 77 c. A random sample of size 219 yielding a sample mean of less than 74.2Incorrect. The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probability of each of the following occurring from this population. Appendix A Statistical Tables a. A random sample of size 36 yielding a sample mean of 78 or more b. A random sample of size 150 yielding a sample mean of between 71 and 77 c. A random sample of size 219 yielding a sample mean of less than 74.2

Answers

Answer 1

Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 p-value, 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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

The mean of a population is 74 and the standard deviation is 15.

This means that [tex]\mu = 74, \sigma = 15[/tex]

Question a:

Sample of 36 means that [tex]n = 36, s = \frac{15}{\sqrt{36}} = 2.5[/tex]

This probability is 1 subtracted by the pvalue of Z when X = 78. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{78 - 74}{2.5}[/tex]

[tex]Z = 1.6[/tex]

[tex]Z = 1.6[/tex] has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

Question b:

Sample of 150 means that [tex]n = 150, s = \frac{15}{\sqrt{150}} = 1.2247[/tex]

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{77 - 74}{1.2274}[/tex]

[tex]Z = 2.45[/tex]

[tex]Z = 2.45[/tex] has a pvalue of 0.9929

X = 71

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{71 - 74}{1.2274}[/tex]

[tex]Z = -2.45[/tex]

[tex]Z = -2.45[/tex] has a pvalue of 0.0071

0.9929 - 0.0071 = 0.9858

0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c. A random sample of size 219 yielding a sample mean of less than 74.2

Sample size of 219 means that [tex]n = 219, s = \frac{15}{\sqrt{219}} = 1.0136[/tex]

This probability is the pvalue of Z when X = 74.2. So

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{74.2 - 74}{1.0136}[/tex]

[tex]Z = 0.2[/tex]

[tex]Z = 0.2[/tex] has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2


Related Questions

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Answers

Answer:

9.24 m to 2 DP's.

Step-by-step explanation:

Consider the right triangle formed from a side , altitude and 1/2 the base.

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x^2 = 8^2 + (0.5x)^2     where x is the length of each side.

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

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Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Step-by-step explanation:

Step 1: Define

y = 4x - 10

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Step 2: Evaluate

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:


At the market Sydney put a few pairs in a bag then ways the bag on a product scale the bag pairs way 4.75 pounds she adds more pairs to the bag the new way of the bag is now 5.6 pounds right and solve an addition equation to find the weight of the pairs added to the bag?

Answers

Answer:

0.85 pounds

Step-by-step explanation:

We start with a bag that weighs 4.75 pounds and want to find the weight of the pairs added to the bag.

Let x equal the weight of the pairs added to the bag.

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Answers

Answer:

177

Step-by-step explanation:

She burned 177 calories

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Answers

Answer:

5*10.88 + 5*11.87=113.75

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Step-by-step explanation:

The perimeter of a square of side 5 cm is 25 cm.
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Answers

Answer:

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Step-by-step explanation:

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Answer:

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in

Answers

Answer:

172.27 in

Step-by-step explanation:

Apply the Sine Rule

Sine Rule is given as:

[tex] \frac{a}{sin(A)} = \frac{b}{sin(B)} = \frac{c}{sin(C)} [/tex]

c = x = ?

C = 180 - (52 + 48) = 80° (sum of triangle)

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Answers

Answer:

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Step-by-step explanation:

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Answers

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When we have an interval with lower bound m and upper bound M, such that:

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Answers

Answer:

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Answers

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Answers

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Step-by-step explanation:

Given data :

currency exchange rate : 1 AUD = 5.8 HKD

cost of each ounce = 2 AUD

Fixed shipping cost for each carton = 80 HKD

number of cartons = 20

next determine the total cost of the 20 cartons in HKD

= ∑(weight in ounce * cost of each ounce *exchange rate) +fixed shipping cost

= ∑ ( 160*2*5.8 + 80 )  + -------------- + (650 *2*5.8 + 80 ) ----------------- ( 1 )

= 81756 HKD

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= 81756 / 20 = 4087.8

ii) Find the standard deviation

= [tex]\sqrt{\frac{(1936-4087.8)^2+-----+(7620-4087.8)^2}{20-1} }[/tex]      note: std =  √∑(xi-X )^2 / (n-1)

= 1933.1817

iii) Find the range

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iv) Determine the third quartile

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Step-by-step explanation:

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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he FDAL level for insect filth in chocolate is 0.6 insect fragments (larvae, eggs, body parts, and so on) per grams. Suppose that there is an average of 0.57 insect fragments per gram of chocolate bar and the distribution is following a Poisson distribution. Compute the probabilities that the number of insect fragments in a 50-gram sample of chocolate bar are:

Answers

Answer:

Step-by-step explanation:

[tex]\text{To find :} \\ \\ 1.) \ \ \text{More than 20} \\ \\ 2) \text{Less than 30 } \\ \\ 3) \ \text{Within the range of }( \mu - \sigma , \mu +\sigma)[/tex]

[tex]\ \ \ Since \ \ X \ \text{follows a poisson distribution and;} \\ \\ X\sim Poisson (0.57) per \ gram \\ \\ X \sim Poisson (28.5) \ per \ 50 \ grams \\ \\[/tex]

[tex]1) \ \ \ \text{The probability } P(X>20) = \sum \limits ^{50}_{x=21} \dfrac{e^{-28.5}\times 28.5^x}{x!} \\ \\ \mathbf{ = 0.9389}[/tex]

[tex]2) P( X<30) = \sum \limits ^{29}_{x=0} \dfrac{e^{-28.5}\times 28.5^x}{x!} \\ \\ \mathbf{=0.5861}[/tex]

[tex]\text{3) Since then mean}\ \mu = 28.5 \\ \\ \sigma =\sqrt{28.5} = 5.339 \\ \\ \mu - \sigma = 28.5 -5.339= 23.161 \\ \\ \mu + \sigma = 28.5 + 5.339 = 33.839[/tex]

[tex]P(23.161 < X< 33.839) = \sum \limits ^{33}_{24} \dfrac{e^{-28.5}\times 28.5^x}{x! } \\ \\ = P(X\le 33) - P( X \le 23) \\ \\ = 0.82678 - 0.1750 \\ \\ \mathbf{= 0.6516}[/tex]

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Answers

Answer:

24ounces

Step-by-step explanation:

16 ounces is 2x 8

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2:3 = 16:24

A person places $277 in an investment account earning an annual rate of 6.7%,
compounded continuously. Using the formula V = Pem, where V is the value of the
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logarithm, and r is the rate of interest, determine the amount of money, to the
nearest cent, in the account after 10 years.

Answers

9514 1404 393

Answer:

  $541.32

Step-by-step explanation:

Putting the given values into the given formula, you have ...

  V = P·e^(rt)

  V = $277e^(0.067·10) ≈ $541.32

ANSWER AND EXPLAIN I’LL MARK BRAINLIEST PLEASE HELP

Answers

Answer:

Im sorry I cant see the link.

Step-by-step explanation:

Answer:

x = 2

y = 1

Step-by-step explanation:

The explain in the above photo

Sorry for the little words, i will rewrite the last two rows in the comment

I hope that is useful for you

3. If an investment company pays 6% compounded semiannually, how much should you deposit
now to have $10,000 5 years from now?

Answers

Answer:

$9,287.67

Step-by-step explanation:

Number of years 5

Interest per pd 3%

FV 10,000

Calculator.net has a financial calculator that will greatly help you

The formula for the amount earned with compound interest after n years is given as:

A = P [tex](1 + r/n)^{nt}[/tex]

P = principal

R = rate

t = time in years

n = number of times compounded in a year.

The amount to deposits so that after 5 years, 10,000 is the amount is

$7,440.94

What is compound interest?

It is the interest we earned on the interest.

The formula for the amount earned with compound interest after n years is given as:

A = P [tex](1 + r/n)^{nt}[/tex]

P = principal

R = rate

t = time in years

n = number of times compounded in a year.

We have,

r = 6%

n = 2

A = $10,000

t = 5 years

A = P [tex](1 + r/n)^{nt}[/tex]

P = 10,000 / [tex](1 + 0.03)^{10}[/tex]

P = 10,000 / [tex](1.03)^{10}[/tex]

P = $7,440.94

Thus,

The amount to deposits so that after 5 years, 10,000 is the amount is

$7,440.94

Learn more about compound interest here:

https://brainly.com/question/13155407

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Answers

Answer:

[tex] 9\frac{1}{3} [/tex]

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Answers

Answer:

x=-36

Step-by-step explanation:

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