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tigry1 [53]
3 years ago
9

The length of time a person takes to decide which shoes to purchase is normally distributed with a mean of 8.54 minutes and a st

andard deviation of 1.91. Find the probability that a randomly selected individual will take less than 5 minutes to select a shoe purchase. Is this outcome unusual?
Mathematics
1 answer:
Minchanka [31]3 years ago
8 0

Answer:

0.0323 = 3.23%. Is unusual

Step-by-step explanation:

Using the normal distribution we can find z value with the mean and standard deviation as follows. x is the value we want to know its probability

z = (x - mean)/standard deviation

z = (5-8.54)/1.91

z = -1.85

Using z tables for normal distribution, for z = 1.85, we have an area under the curve of 0.9677. Since we want to know the probability for 5 minutes or less, we have to substract from

p = 1- 0.9677

p = 0.0323

An event with a probability less than 5% is considered unusual.

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Answer:

No, a porportional relationship is not shown

Step-by-step explanation:

Up until 4(x) and 10(y), the relationship is porportional. Then, when x increases by 2, y also only increases by 2. Y would need to increase by 5.

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Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. Whe
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Answer:

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

n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

And rounded up we have that n=1068

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".  

The population proportion have the following distribution

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

Solution to the problem

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 0.03 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 don't have a prior estimation for the proportion \hat p so we can use 0.5 as an approximation for this case  

And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

And rounded up we have that n=1068

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