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BaLLatris [955]

2 years ago

5

chris has a bag of sweets. there are more than 20 sweets in the bag he shares his sweets equally between six people chris says,

"the number of sweets in the packet was prime." is he correct? circle: yes or no, explain your answer below.

1 answer:

Likurg_2 [28]2 years ago

3 0

No, this is because a prime number is an uneven number. So it can not be prime.

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It is believed that as many as 23% of adults over 50 never graduated from high school. We wish to see if this percentage is the

JulijaS [17]

Answer:

1) n=48

2) n=298

3) n=426

Step-by-step explanation:

Previous concepts

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 means 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.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p$ represent the real population proportion of interest

$\hat p$ represent the estimated proportion for the sample

n is the sample size required (variable of interest)

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

Part 1

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.10$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.1$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:

$n=\frac{0.23(1-0.23)}{(\frac{0.1}{1.64})^2}=47.63$

And rounded up we have that n=48

Part 2

The margin of error on this case changes to 0.04 so if we use the same formula but changing the value for ME we got:

$n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.64})^2}=297.7$

And rounded up we have that n=298

Part 3

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.04$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:

$n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.96})^2}=425.22$

And rounded up we have that n=426

3 0

2 years ago

A sample of 1714 cultures from individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiot

Damm [24]

Answer:

a) $p$ represent the real population proportion of people who showed partial or complete resistance to the antibiotic

b) $\hat p=\frac{973}{1714}=0.568$ represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

c) We are confident (95%) that the true proportion of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic is between 54.5% to 59.1%.

Step-by-step explanation:

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 means 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.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Part a

Description in words of the parameter p

$p$ represent the real population proportion of people who showed partial or complete resistance to the antibiotic

$\hat p$ represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

n=1714 is the sample size required

$z_{\alpha/2}$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Part b

Numerical estimate for p

In order to estimate a proportion we use this formula:

$\hat p =\frac{X}{n}$ where X represent the number of people with a characteristic and n the total sample size selected.

$\hat p=\frac{973}{1714}=0.568$ represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

Part c

The confidence interval for a proportion is given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.568 - 1.96 \sqrt{\frac{0.568(1-0.568)}{1714}}=0.545$

$0.568 + 1.96 \sqrt{\frac{0.56(1-0.56)}{1714}}=0.591$

And the 95% confidence interval would be given (0.545;0.591).

We are confident that about 54.5% to 59.1% of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic.

5 0

1 year ago

Barbara drives between Miami, Florida, and West Palm Beach, Florida. She drives 50 mi in clear weather and then encounters a thu

NeX [460]

Distance traveled in clear weather = 50 miles

Distance traveled in thunderstorm = 15 miles

Let speed in clear weather = x

⇒ Speed in thunderstorm = x-20

Total time taken for trip = 1.5 hours

We need to determine average speed in clear weather (i.e. x) and average speed in the thunderstorm (i.e. x-20 ).

Total time taken for trip = Time taken in clear weather + Time taken in thunderstorm

⇒ Total time taken for trip = $\frac{Distance covered in clear weather}{Speed in clear weather}$ + $\frac{Distance covered in thunderstorm}{Speed in thunderstorm}$

⇒ 1.5 = $\frac{50}{x}$ + $\frac{15}{x-20}$

⇒ 1.5 = $\frac{50(x-20)+15(x)}{(x)(x-20)}$

⇒ 15*x*(x-20) = 10*[50*(x-20)+15*x]

⇒ 15x² - 300x = 500x - 10,000 + 150x

⇒ 15x² - 300x = 650x - 10,000

⇒ 15x² - 950x + 10,000 = 0

⇒ 3x² - 190x + 2,000 = 0

The above equation is in the format of ax² + bx + c = 0

To determine the roots of the equation, we will first determine 'D'

D = b² - 4ac

⇒ D = (-190)² - 4*3*2,000

⇒ D = 36,100 - 24,000

⇒ D = 12,100

Now using the D to determine the two roots of the equation

Roots are: x₁ = $\frac{-b+\sqrt{D}}{2a}$ ; x₂ = $\frac{-b-\sqrt{D}}{2a}$

⇒ x₁ = $\frac{-(-190)+\sqrt{12,100}}{2*3}$ and x₂ = $\frac{-(-190)-\sqrt{12,100}}{2*3}$

⇒ x₁ = $\frac{190+110}{6}$ and x₂ = $\frac{190-110}{6}$

⇒ x₁ = $\frac{300}{6}$ and x₂ = $\frac{80}{6}$

⇒ x₁ = 50 and x₂ = 13.33

So speed in clear weather can be 50 mph or 13.33 mph. However, we know that in thunderstorm was 20 mph less than speed in clear weather.

If speed in clear weather is 13.33 mph then speed in thunderstorm would be negative, which is not possible since speed can't be negative.

Hence, the speed in clear weather would be 50 mph, and in thunderstorm would be 20 mph less, i.e. 30 mph.

7 0

2 years ago

Cuando un oso grizzly hiberna, su ritmo cardiaco disminuye a 101010 latidos por minuto, que es 20\%20%20, percent de su valor no

zheka24 [161]

Answer:

La frecuencia cardíaca normal de un oso pardo cuando no está hibernando es de 50 latidos por minuto.

Step-by-step explanation:

5 0

1 year ago

Sixty-eight percent of adults in a certain country believe that life on other planets is plausible. You randomly select five adu

Pavlova-9 [17]

Answer:

Mean = 3.4

Variance = 1.088

Standard deviation = 1.0431

Step-by-step explanation:

$P=0.68\\ \\N=5$

The mean of a binomial distribution with parameters N (the number of trials) and p (the probability of success for each trial) is $m=N\cdot p$

Thus,

$m=5\cdot 0.68=3.4$

The variance of the binomial distribution is $s^2=Np(1-p)$ , where $s^2$ is the variance of the binomial distribution, so

$s^2=5\cdot 0.68\cdot (1-0.68)=3.4\cdot 0.32=1.088$

The standard deviation $s$ is the square root of the variance $s^2,$ so

$s=\sqrt{1.088}\approx 1.0431$

3 0

1 year ago