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Nonamiya [84]
2 years ago
6

An advertiser drops 10,000 leaflets on a city which has 2000 blocks. Assume that each leaflet has an equal chance of landing on

each block. What is the probability that a particular block will receive no leaflets
Mathematics
1 answer:
Ira Lisetskai [31]2 years ago
5 0

Answer:

0.00673

Step-by-step explanation:

As each of the 10000 leaflets has equal chance on landing into 1 of the 2000 blocks, this means each leaflet has a probability of 1/2000 to land on the 1st block, 1/2000 to land on the 2nd lock, 1/2000 to land on the 3rd block, and so on.

The probability that a particular block will receive no leaflet is chance of each leaflets to land on the other 1999 blocks, which has a chance of 1999/2000. For all 10000 leaflets to happen like this independently, then the total probability is

\left(\frac{1999}{2000}\right)^{10000} = 0.00673

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Answer:

95% Confidence Interval: 4.7 ± 1.57

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Amir lent Rs. 600 to Hameed for 2 yrs and Rs.150 to Aman for 4 yrs and recievd altogether from both rs.90 as simple intrest.The
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1 year ago
Read 2 more answers
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
Grandma is trying out a new recipe for raisin bread. Each batch of bread dough makes three loaves, and each loaf contains 15 sli
Tom [10]

Answer:

0.108

Step-by-step explanation:

Using the poisson probability process :

Where :

P(x =x) = (e^-λ * λ^x) ÷ x!

Given that :

Each batch of bread = 3 loaves

Each loaf = 15 slices

Total slice per batch = 15 * 3 = 45 slices

Number of raising added = 100

Average number of raisin per slice, λ = 100/45 = 20/9

Hence,

Probability that a randomly chosen slice has no raising :

P(x = 0) = (e^-λ * λ^x) ÷ x!

P(x = 0) = (e^-(100/45) * (100/45)^0) ÷ 0!

P(x = 0) = (0.1083680 * 1) / 1

P(x = 0) = 0.108

8 0
2 years ago
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