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SVETLANKA909090 [29]

2 years ago

10

To be able to say with 95% confidence level that the standard deviation of a data set is within 10% of the population's standard

deviation, the number of observations within the data set must be greater than or equal to what quantity?

1 answer:

boyakko [2]2 years ago

8 0

The answer in this question is 97.
0.20 SD = 1.96 SD / sqrt(n)
n = (1.96 / .200)^2
n = 96.04
Which is rounded Up to 97
The number of observations within the data set must be greater than or equal to the quantity of 97.

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Which point is an x-intercept of the quadratic function f(x)=(x-8)(x-9)?

Nonamiya [84]

The zeroes are (0,8) and (0,9)

Use the zero product property to find this
(x-8)=0
x=8

(x-9)=0
x=9

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2 years ago

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A construction company is considering submitting bids for contracts of three different projects. The company estimates that it h

julsineya [31]

Answer:

a. $P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\$

b. $E(x) = 0.3$

c. $S(x)=0.5196$

d. $E=5,000$

Step-by-step explanation:

The probability that the company won x bids follows a binomial distribution because we have n identical and independent experiments with a probability p of success and (1-p) of fail.

So, the PMF of X is equal to:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\$

Where p is 0.1 and it is the chance of winning. Additionally, n is 3 and it is the number of bids. So the PMF of X is:

$P(x)=\frac{3!}{x!(3-x)!}*0.1^{x}*(1-0.1)^{n-x}\\$

For binomial distribution:

$E(x)=np\\S(x)=\sqrt{np(1-p)}$

Therefore, the company can expect to win 0.3 bids and it is calculated as:

$E(x) = np = 3*0.1 = 0.3$

Additionally, the standard deviation of the number of bids won is:

$S(x)=\sqrt{np(1-p)}=\sqrt{3(0.1)(1-0.1)}=0.5196$

Finally, the probability to won 1, 2 or 3 bids is equal to:

$P(1)=\frac{3!}{1!(3-1)!}*0.1^{1}*(1-0.1)^{3-1}=0.243\\P(2)=\frac{3!}{2!(3-2)!}*0.1^{2}*(1-0.1)^{3-2}=0.027\\P(3)=\frac{3!}{3!(3-3)!}*0.1^{3}*(1-0.1)^{3-3}=0.001$

So, the expected profit for the company is equal to:

$E=-10,000+50,000(0.243)+100,000(0.027)+150,000(0.001)\\E=5,000$

Because there is a probability of 0.243 to win one bid and it will produce 50,000 of income, there is a probability of 0.027 to win 2 bids and it will produce 100,000 of income and there is a probability of 0.001 to win 3 bids and it will produce 150,000 of income.

5 0

2 years ago

Choose all sequences of transformations that produce the same image of a given figure.

elena-14-01-66 [18.8K]

<em><u>Your answer: </u></em> a reflection across the y-axis followed by a clockwise rotation 90° about the origin. A clockwise rotation 90° about the origin followed by a reflection across the x-axis. A counter-clockwise rotation 90° about the origin followed by a reflection across the y-axis. A reflection across the x-axis followed by a counter-clockwise rotation 90° about the origin.

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2 years ago

A store had 4 boxes of video Games.How many days would it take to sell the games if each day they sold one fifth

Korvikt [17]

The answer would be 20 days.

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2 years ago

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According to a survey, 10% of Americans are afraid to fly. Suppose 1100 Americans are sampled. a) What is the probability that 1

JulijaS [17]

Answer:

a) 14.46%

b) 0.00%

c) 1.54%

Step-by-step explanation:

According to the survey, 10% of Americans are afraid to fly.

This means $p=0.10$ and $q=1-0.10=0.90$ .

If 1100 Americans are sampled, then the sample size is $n=1100$ .

The mean of the distribution is $\mu=np$ .

This means $\mu=1100\times 0.10=110$

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

We substitute the values to get:

$\sigma=\sqrt{1100\times 0.1\times 0.9}=9.95$

a) We want to find the probability that 121 or more Americans in the survey are afraid to fly.

We first apply the continuity correction factor to get: $P(X\ge121)=P(X\:>\:121-0.5)\\P(X\ge121)=P(X\:>\:120.5)$

We now convert to Z-scores to get:

$P(X\:>\:120.5)=P(z\:>\:\frac{120.5-110}{9.95})=1.06\\$

From the standard normal distribution table P(z>1.06)=0.1446

As a percentage, the probability is 14.46%

b) We want to find the probability that 165 or more Americans in the survey are afraid to fly.

We apply the CCF to get:

$P(X\ge165)=P(X\:>\:165-0.5)\\P(X\ge165)=P(X\:>\:164.5)$

We convert to z-scores:

$P(X\:>\:164.5)=P(z\:>\:\frac{164.5-110}{9.95})=5.48$

From the normal distribution, P(z>164.5)=0

c) First, 8% of 1100 is 88.

We want to find the probability that 88 or less Americans in the survey are afraid to fly.

We apply the CCF to get:

$P(X\le88)=P(X\:$

We convert to z-scores:

$P(X\:$

From the normal distribution, P(z>\:-2.16)=0.0154

As a percentage, we get 1.54%

3 0

2 years ago