Based on a​ poll, 67​% of Internet users are more careful about personal information when using a public​ Wi-Fi hotspot. What is the probability that among four randomly selected Internet​ users, at least one is more careful about personal information when using a public​ Wi-Fi hotspot? How is the result affected by the additional information that the survey subjects volunteered to​ respond?

Answer:

Step-by-step explanation:

1) The z value was determined using a normal distribution table. From the normal distribution table, the corresponding z value for a 94% confidence interval is 1.88

The correct option is D

2) if 70% of all internet users experience e-mail fraud. It means that probability of success, p

p = 70/100 = 0.7

q = 1 - p = 1 - 0.7 = 0.3

n = number of selected users = 50

Mean, u = np = 50×0.7 = 35

Standard deviation, u = √npq = √50×0.7×0.3 = 3.24

x = number of internet users

The formula for normal distribution is expressed as

z = (x - u)/s

We want to determine the probability that no more than 25 were victims of e-mail fraud. It is expressed as

P(x lesser than or equal to 25)

The z value will be

z = (25- 35)/3.24 = - 10/3.24 = -3.09

Looking at the normal distribution table, the corresponding z score is 0.001

P(x lesser than or equal to 25) = 0.001

Answer:

[tex]z=2.53[/tex]  

[tex]p_v =P(z>2.53)=0.0057[/tex]  

The p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so then we have enough evidence to reject the null hypothesis, and we can say that at 5% of significance, the proportion of adults who support a ban on sugary snacks and soft​ drinks  is more than 0.5 or 50%.

Step-by-step explanation:

1) Data given and notation

n=1000 represent the random sample taken

X=540 represent the adults that said that schools should ban sugary snacks and soft drinks

[tex]\hat p=\frac{540}{1000}=0.54[/tex] estimated proportion of adults that said that schools should ban sugary snacks and soft drinks

[tex]p_o=0.5[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that majority of adults​ (more than​ 50%) support a ban on sugary snacks and soft​ drinks, the system of hypothesis are:  

Null hypothesis:[tex]p\leq 0.5[/tex]  

Alternative hypothesis:[tex]p > 0.5[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.54 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.53[/tex]  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a one side right tailed test the p value would be:  

[tex]p_v =P(z>2.53)=0.0057[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so then we have enough evidence to reject the null hypothesis, and we can say that at 5% of significance, the proportion of adults who support a ban on sugary snacks and soft​ drinks  is more than 0.5 or 50%.

Answer:

The 98% confidence interval for the mean repair cost for the dryers is (83.4161, 103.3039).

Step-by-step explanation:

We have a small sample size n = 25, [tex]\bar{x} = 93.36[/tex] and s = 19.95. The confidence interval is given by  [tex]\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})[/tex] where [tex]t_{\alpha/2}[/tex] is the [tex]\alpha/2[/tex]th quantile of the t distribution with n - 1 = 25 - 1 = 24 degrees of freedom. As we want the 98% confidence interval, we have that [tex]\alpha = 0.02[/tex] and the confidence interval is [tex]93.36\pm t_{0.01}(\frac{19.95}{\sqrt{25}})[/tex] where [tex]t_{0.01}[/tex] is the 1st quantile of the t distribution with 24 df, i.e., [tex]t_{0.01} = -2.4922[/tex]. Then, we have [tex]93.36\pm (-2.4922)(\frac{19.95}{\sqrt{25}})[/tex] and the 98% confidence interval is given by (83.4161, 103.3039).

Answer:

Step-by-step explanation:

Our aim is to determine a 98% confidence interval for the mean repair cost for the dryers

Number of samples. n = 25

Mean, u = $93.36

Standard deviation, s = $19.95

For a confidence level of 98%, the corresponding z value is 2.33. This is determined from the normal distribution table.

We will apply the formula,

Confidence interval

= mean ± z × standard deviation/√n

It becomes

93.36 ± 2.33 × 19.95/√25

= 93.36 ± 2.33 × 3.99

= 93.36 ± 9.2967

The lower boundary of the confidence interval is 93.36 - 9.2967 =84.0633

The upper boundary of the confidence interval is 93.36 + 9.2967 = 102.6567

Therefore, with 98% confidence interval, the mean repair costs for the dryers is between $84.0633 and $102.6567

Answer: option A is correct

Step-by-step explanation:

the prediction interval is to cover a “moving target”, the random future value of y,

while the confidence interval is to cover the “fixed target”,

the average (expected) value of y, E(y).

Answer:

69.5%

Step-by-step explanation:

A feature of the normal distribution is that this is completely determined by its mean and standard deviation, therefore, if two normal curves have the same mean and standard deviation we can be sure that they are the same normal curve. Then, the probability of getting a value of the normally distributed variable between 6 and 8 is 0.695. In practice we can say that if we get a large sample of observations of the variable, then, the percentage of all possible observations of the variable that lie between 6 and 8 is 100(0.695)% = 69.5%.

Answer:

  A. $1,510.10

Step-by-step explanation:

Fill in the given numbers and do the arithmetic.

  M = I·(1+r)^n

  M = $1423.00·(1 +.02)^3 = $1510.099

  M ≈ $1510.10

Answer:

We conclude that the lifetime of tires is less than 30,000 miles.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 30,000 miles

Sample mean, [tex]\bar{x}[/tex] = 29,400 miles

Sample size, n = 54

Alpha, α = 0.05

Population standard deviation, σ =1200 miles

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 30000\text{ miles}\\H_A: \mu < 30000\text{ miles}[/tex]

We use One-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{29400 - 30000}{\frac{1200}{\sqrt{54}} } = -3.6742[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = -1.64[/tex]

Since,  

[tex]z_{stat} < z_{critical}[/tex]

We reject the null hypothesis and accept the alternate hypothesis.

Thus, we conclude that the lifetime of tires is less than 30,000 miles.

Answer:

The 98% confidence interval for the variance in the pounds of impurities would be [tex]8.400 \leq \sigma^2 \leq 39.827[/tex].

Step-by-step explanation:

1) Data given and notation

[tex]s^2 =16[/tex] represent the sample variance

s=4 represent the sample standard deviation

[tex]\bar x[/tex] represent the sample mean

n=20 the sample size

Confidence=98% or 0.98

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

[tex]df=n-1=20-1=19[/tex]

Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.01,19)" "=CHISQ.INV(0.99,19)". so for this case the critical values are:

[tex]\chi^2_{\alpha/2}=36.191[/tex]

[tex]\chi^2_{1- \alpha/2}=7.633[/tex]

And replacing into the formula for the interval we got:

[tex]\frac{(19)(16)}{36.191} \leq \sigma^2 \leq \frac{(19)(16)}{7.633}[/tex]

[tex] 8.400 \leq \sigma^2 \leq 39.827[/tex]

So the 98% confidence interval for the variance in the pounds of impurities would be [tex]8.400 \leq \sigma^2 \leq 39.827[/tex].

The 98% confidence interval for the mean change in score is between 69.92 points to 74.08 points

Standard deviation = √variance = √16 = 4

The z score of  98% confidence interval is 2.326

The margin of error (E) is:

[tex]E = Z_\frac{\alpha }{2} *\frac{standard\ deviation}{\sqrt{sample\ size} } =2.326*\frac{4}{\sqrt{20} } =2.08[/tex]

The confidence interval = mean ± E = 72 ± 2.08 = (69.92, 74.08)

The 98% confidence interval for the mean change in score is between 69.92 points to 74.08 points

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Answer:

Step-by-step explanation:

Hello!

You have X~Bi (n;ρ)

Where:

n=10

ρ= 0.10

For all asked probabilities you need to use a Binomial distribution table. Remember this table has the information of the cummulative probabilities P(X≤x).

a. f(0) ⇒ P(X=0) = 0.3487

b. f(2) ⇒ P(X=2) ⇒ P(X≤2) - P(X≤1) = 0.9298 - 0.7361 = 0.1937

c. P(X≤2) = 0.9298

d. P(X ≥ 1) = 1 - P(X ≤ 1) = 1 - 0.7361 = 0.2639

e. E(X)= nρ = 10*0.10 = 1

f. V(X)= nρ(1-ρ) = 10*0.1*0.9 = 0.9

σ= √V(X) = √0.9 = 0.9487

I hope it helps!

The binomial experiment depicts that f(0) will be 0.3487.

From the information given, it can be noted that:

n = 10

p = 0.10

q = 1 - 0.10 = 0.90

f(0) ⇒ P(X=0) = 0.3487

f(2) = P(X=2) ⇒ P(X≤2) - P(X≤1)

= 0.9298 - 0.7361

= 0.1937

P(X≤2) = 0.9298

P(X ≥ 1) = 1 - P(X ≤ 1)

= 1 - 0.7361 = 0.2639

E(X) = nρ

= 10 × 0.10 = 1

V(X) = nρ(1-ρ)

= 10 × 0.1 × 0.9

= 0.9

σ = √V(X) = √0.9

= 0.9487

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Answer:

Step-by-step explanation:

To find the probability that the average tax paid on the sample forms is greater than $1980, standardize the sample mean using the Central Limit Theorem. To find the probability that more than 60 of the sampled forms have a tax greater than $3200, use a normal distribution approximation.

To solve this problem, we can use the Central Limit Theorem.

a. To find the probability that the average tax paid on the sample forms is greater than $1980, we need to standardize the sample mean. We calculate the z-score by subtracting the population mean from the sample mean and dividing by the population standard deviation divided by the square root of the sample size. Then, we can use a z-table or a calculator to find the probability.

b. To find the probability that more than 60 of the sampled forms have a tax greater than $3200, we can use a normal distribution approximation. We can calculate the z-score for 60 forms by subtracting the population mean from 60 and dividing by the square root of the population mean multiplied by (1 - the population mean) divided by the sample size. Then, we can use a z-table or a calculator to find the probability.

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Answer:

option 0.8%

Step-by-step explanation:

Data provided in the question:

Mean = 5.7 years

Standard deviation, s = 1.8 years

Now,

P(the employee has worked at the store for over 10 years)

= P(X > 10 years)

= [tex]P (Z > \frac{X-Mean}{\sigma})[/tex]

or

= [tex]P (Z > \frac{10-5.7}{1.8})[/tex]

= P (Z > 2.389 )

or

= 0.008447     [from standard  z table]

or

= 0.008447 × 100% = 0.84% ≈ 0.8%

Hence,

the correct answer is option 0.8%

To find the probability that an employee has worked at a large department store for over 10 years, we can use the z-score formula and a standard normal distribution table or calculator.

To find the probability that an employee has worked at the store for over 10 years, we can use the z-score formula:

z = (x - μ) / σ

where x is the value we are interested in (10 years), μ is the mean (5.7 years), and σ is the standard deviation (1.8 years).

Plugging in the values, z = (10 - 5.7) / 1.8 = 2.39

Using a standard normal distribution table or a calculator, we can find that the probability of a z-score greater than 2.39 is approximately 0.008, or 0.8%.

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Answer:

skew lines

Step-by-step explanation:

we are given 2 lines in parametric form as

L1:x=18+6t,y=7+3t,z=13+3t and L2:x=−14+7ty=−12+5tz=−8+6t[tex]L1:x=18+6t,y=7+3t,z=13+3t \\ L2:x=-14+7t,y=-12+5t,z=-8+6t[/tex]

If the lines intersect then the two points must be equal for one value of t.

Let us try equating x,y and z coordinate.

[tex]18+6t = -14+7t\\t=32\\[/tex]

when we equate y coordinate we get

[tex]7+3t =-12+5t\\2t =19\\t =9.5[/tex]

Since we get two different t we find that these two lines cannot intersect.

Comparing direction ratios we have

I line has direction ratios as (6,3,3) and second line (7,5,6)

These two are not proportional and hence not parallel

So these lines are skew lines

After analyzing the direction vectors of lines L1 and L2, it is clear that they are neither parallel nor do they intersect, which means the lines are skew.

To determine whether the lines L1 and L2 intersect, are skew, or are parallel, we need to compare the direction vectors and check if they can be expressed as multiples of each other. If we consider the components of each direction vector given by the 't' terms in the equations of L1 and L2, they are (6, 3, 3) for L1 and (7, 5, 6) for L2 respectively.

To check for parallelism, we find a constant 'k' such that (6, 3, 3) = k*(7, 5, 6). After trying to solve for 'k', we immediately see that no such constant can exist, as 6/7 is not equal to 3/5 or 3/6. Therefore, these lines are not parallel.

To check for intersection, we need to find a common point that satisfies both line equations for some value of the parameter 't'. However, since an attempt to find such values results in inconsistent equations, it suggests that no such point exists and these lines do not intersect.

Given that the lines are neither parallel nor do they intersect, they must be skew lines. Skew lines are lines that do not intersect and are not parallel, lying in different planes.

Answer: f(-) = - 12

Step-by-step explanation:

The function is expressed as

f(x)= -x^2+10x-3

To determine f(-1), we will substitute

x = - 1 into the given function. It becomes

f(-1) = (- 1)^2 + 10(-1) - 3

f(-1) = 1 - 10 - 3

f(-1) = - 12

Answer:

It was originally $91.25

Answer:

20 inches

Step-by-step explanation:

Take 2/3 of an inch 30 times.

The height of the model is then 30 times 2/3 inches = 10*2 = 20 inches

Answer:

2/3=0.6667

Step-by-step explanation:

Let X be the cycle time for  trucks hauling concrete to a highway construction site

Given that X is U(50,70)

Hence pdf of X is

[tex]f(x) = \frac{1}{20} ,50<x<70[/tex]

Let A be the event that cycle time is no more than 65 minutes and

B the event cycle time exceeds 55 minutes

Required probability

=  the conditional probability that the cycle time is no more than 65 minutes if it is known that the cycle time exceeds 55 minutes

= P(A/B)

=[tex]\frac{P(A\bigcapB)}{P(B)} \\=\frac{P(55<x<65)}{P(X>55)} \\=\frac{65-55}{70-55} \\=\frac{2}{3}[/tex]

Answer:

He can make the same decision at a = 0.10, but he may not make the same decision at a = 0.01

Step-by-step explanation:

In hypothesis tests, critical regions are ranges of the distributions where the values represent statistically significant results. Analysts define the size and location of the critical regions by specifying both the significance level (alpha) and whether the test is one-tailed or two-tailed.

As the significance level gets bigger, the range of the critical region increases.

Therefore significant results at lower significance levels are still significant at higher significance levels. Thus significant result at a=0.05 is always significant at a=0.10

But  significant result at a=0.05 may not be significant at a=0.01 since critical region shrinks, therefore the result may not fall in the critical region at a=0.01

The P(69 ≤ X ≤ 71) when n = 16 is approximately 0.9312.

Here's the calculation for P(69 ≤ X ≤ 71) when n = 16:

1. Find the standard error (SE):

SE = σ / √n = 2.2 GPa / √16 = 0.55 GPa

2. Standardize the values:

Z1 = (69 - 70) / 0.55 = -1.82

Z2 = (71 - 70) / 0.55 = 1.82

3. Use a standard normal table or calculator to find the probabilities:

P(Z ≤ 1.82) = 0.9656

P(Z ≤ -1.82) = 0.0344

4. Calculate the probability of the interval:

P(69 ≤ X ≤ 71) = P(-1.82 ≤ Z ≤ 1.82) = P(Z ≤ 1.82) - P(Z ≤ -1.82)

= 0.9656 - 0.0344 = 0.9312

Therefore, P(69 ≤ X ≤ 71) when n = 16 is approximately 0.9312.

Answer:

8 or 3

Step-by-step explanation:

Answer:

Yes, the events T1 and T2 are independent.

Step-by-step explanation:

When flipping a fair coin, for each flip, there is a 50/50 chance that it will result in heads or tails.

For the first flip, P(T1) = 0.5

For the second flip, P(T2) = 0.5 regardless of the outcome of the first flip. Therefore, T1 and T2 are independent events.

*Note that if the question asked for the event of both flips resulting in tails, then the events would be dependent.

Answer:

(a) 1/2

(b) 1/2

(c) 1/8

Step-by-step explanation:

Since, when a fair coin is tossed three times,

The the total number of possible outcomes

n(S) = 2 × 2 × 2

= 8 { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT },

Here, B : { At least two heads are observed } ,

⇒ B = {HHH, HHT, HTH, THH},

⇒ n(B) = 4,

Since,

[tex]\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}[/tex]

(a) So, the probability of B,

[tex]P(B) =\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}[/tex]

(b) A : { At least one head is observed },

⇒ A = {HHH, HHT, HTH, THH, HTT, THT, TTH},

∵ A ∩ B = {HHH, HHT, HTH, THH},

n(A∩ B) = 4,

[tex]\implies P(A\cap B) = \frac{n(A\cap B)}{n(S)} = \frac{4}{8}=\frac{1}{2}[/tex]

(c) C: { The number of heads observed is odd },

⇒ C = { HHH, HTT, THT, TTH},

∵ A ∩ B ∩ C = {HHH},

⇒ n(A ∩ B ∩ C) = 1,

[tex]\implies P(A\cap B\cap C)=\frac{1}{8}[/tex]

Answer:

2, 0.135, 0.270, 0.270, 0.180, 0.145

Step-by-step explanation:

1. To calculate the mean of people arriving in 5 minute period:

We know the arrival rate of minute is 0.4, so people arriving in 5 minutes will be

0.4 x 5 = 2

2. From part 1 it is known thay mean arrival time=2

For this we will use poisson's probability formula that is

P(X=x) = (2^x) x Exp^(-2/x!)

For X=0

P(X=0) = (2^0) x Exp^(-2/0!) = 0.135

For X = 1

P(X=1) = (2^1) x Exp^(-2/1!) = 0.270

For X = 2

P(X=2) = (2^2) x Exp^(-2/2!) = 0.270

For X = 3

P(X=3) = (2^3) x Exp^(-2/3!) = 0.180

3. For delay expected if more than 3 customer arrive in 5 minutes.

P(X>3) = 1 - P(X=0) - P(X=1) - P(X=2) - P(X=3)

P(X>3) = 1-0.135-0.270-0.270-0.180

P(X>3) = 0.145

Answer:

The probability that the mean life a random sample of 25 such blenders falls between 4.6 and 5.1 years is 0.6687.

Step-by-step explanation:

We have given :

The average life a manufacturer's blender is 5 years i.e. [tex]\mu=5[/tex]

The standard deviation is [tex]\sigma=1[/tex]

Number of sample n=25.

To find : The probability that the mean life falls between 4.6 and 5.1 years ?

Solution :

Using z-score formula, [tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

The probability that the mean life falls between 4.6 and 5.1 years is given by, [tex]P(4.6<X<5.1)[/tex]

[tex]P(4.6<X<5.1)=P(\frac{4.6-5}{\frac{1}{\sqrt{25}}}<Z<\frac{5.1-5}{\frac{1}{\sqrt{25}}})[/tex]

[tex]P(4.6<X<5.1)=P(\frac{-0.4}{\frac{1}{5}}<Z<\frac{0.1}{\frac{1}{5}})[/tex]

[tex]P(4.6<X<5.1)=P(-2<Z<0.5)[/tex]

[tex]P(4.6<X<5.1)=P(Z<0.5)-P(Z<-2)[/tex]

Using z-table,

[tex]P(4.6<X<5.1)=0.6915-0.0228[/tex]

[tex]P(4.6<X<5.1)=0.6687[/tex]

Therefore, the probability that the mean life a random sample of 25 such blenders falls between 4.6 and 5.1 years is 0.6687.

The probability that the mean life of a random sample of 25 blenders falls between 4.6 and 5.1 years is approximately 0.3585, or 35.85%.

To find the probability that the mean life of a random sample of 25 blenders falls between 4.6 and 5.1 years, we can use the standard normal distribution. First, we need to standardize the values using the z-score formula. The z-score for 4.6 years is (4.6 - 5) / 1 = -0.4, and the z-score for 5.1 years is (5.1 - 5) / 1 = 0.1. We can then look up the corresponding probabilities in the standard normal distribution table or use a calculator to find the area under the curve between these two z-scores. The probability that the mean life falls in this range is approximately 0.3585, or 35.85%.

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Answer:

D

Step-by-step explanation:

Answer:

(a)C(x) = 2500 + 3x

(b)R(x) = 5.5x

(c)x = 1000

Step-by-step explanation:

(a)His cost function as a function of x

C(x) = 2500 + 3x

(b)His revenue R function as a function of x

R(x) = 5.5x

(c)When revenue R equals to Cost C

C(x) = R(x)

2500 + 3x = 5.5x

2.5x = 2500

x = 1000

(d) Refer to attachment. We can see that the 2 lines intercepts at x = 1000 and y = 2500. That's the point of turning to profit.

Answer:

can you provide the list?

Step-by-step explanation:

Answer: fifth term = -1/16

Sixth term = 1/256

Step-by-step explanation:

The given sequence is a geometric progression. This is because the ratio of two consecutive terms is constant.

We will apply the formula for determining the nth term of a geometric progression series.

Tn = ar^n-1

Where

Tn = value of the nth term of the geometric series

a = The first term of the series.

r = common ratio(ratio of a term to a consecutive previous term)

n = number of terms in the series

From the in information given,

a = -16

r = 4/-16 = -1/4

The next 2 terms are the 5th and 6th terms.

T5 = -16 × -1/4^(5-1)

T5 = -16 × (-1/4)^4

T5 = -16 × 1/256 = -1/16

The 6th term would be the 5th term × the common ratio. It becomes

T6 = -1/16 ×-1/4 = 1/256

For testing if a coin is fair, a binomial test is used to compare the observed heads to what's expected under a fair coin scenario. Differences in p-values for the same sample proportion across different sample sizes are due to the increased precision of larger samples.

In hypothesis testing, when we want to determine whether a coin (or in this case, a hypothetical lizard) is fair, we employ a binomial test. This test assesses the number of 'successes' (heads in our case), comparing it to what we would expect under the null hypothesis of p=0.5, indicating a fair coin.

For part (a), where we get 56 heads out of 100 tosses, we would calculate the p-value by looking at both the proportion of heads obtained and the standard error for a binomial distribution. Although not calculated here directly, if this p-value is less than the chosen significance level, typically 0.05, we reject the null hypothesis, concluding the coin is not fair.

For part (b), getting 560 heads out of 1000 tosses yields the same sample proportion as part (a). However, the larger sample size increases the power of the test, often resulting in a smaller p-value if the observed proportion is different from the expected 0.5.

Comparing p-values from (a) and (b) reveals that despite having the same sample proportions, the p-values differ due to the sample sizes. Larger sample sizes yield more precise estimates of the population proportion, thus potentially resulting in smaller p-values for the same sample proportion deviation from the null hypothesis.

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