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3 years ago

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The probability of an operator entering alphanumeric data incorrectly into a field in a database is equally likely. The random v

ariable X is the number of fields on a data entry form with an error. The data entry form has 28 fields. Is X a discrete uniform random variable? Why or why not? (Please provide the mathematical details that lead to your answer.)

1 answer:

34kurt3 years ago

5 0

Answer:

Since P(0) and P(1) have not the same values for the probability, X is not a discrete uniform random variable.

Step-by-step explanation:

Probability of data to be correct: 0.5

Probability of data being incorrect: 0.5

The random variable X is the number of fields that have an error within the range of 28.

P(0) = $0.5^2^8$

P(1) = $^2^8\text{C}_1 \times 0.5 \times 0.5^2^7 = 28 \times 0.5 ^2^8$

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A retail store did research and determined that, on average, each customer spends 40 minutes in their store per visit, with a st

Blababa [14]

Answer:

0.13% of customers spend more than 46 minutes

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 40, \sigma = 2$

What percentage of customers spend more than 46 minutes?

This is 1 subtracted by the pvalue of Z when X = 46. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{46 - 40}{2}$

$Z = 3$

$Z = 3$ has a pvalue of 0.9987

1 - 0.9987 = 0.0013

0.13% of customers spend more than 46 minutes

6 0

3 years ago

A company is approached by a new warehouse management systems vendor to adopt a new system. The vendor claims that the warehouse

Ludmilka [50]

Answer:

a

The null hypothesis is $H_o : \mu = 180$

The alternative hypothesis is $H_a : \mu < 180$

b

The decision rule is fail to reject the null hypothesis

The conclusion

There no sufficient evidence to support the vendor claim that the warehouse management system reduces the average pick, pack, and ship time to below 3 minutes(180 seconds) per order through bin location and routing optimization

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 3 \ minutes = 180 \ seconds$

The sample size is n = 25

The excel sheet is show on the first uploaded image

The level of significance is $\alpha = 0.05$

The null hypothesis is $H_o : \mu = 180$

The alternative hypothesis is $H_a : \mu < 180$

Generally the sample mean is mathematically represented as

$\= x = \frac{156 + 136 + \cdots + 181}{25}$

=> $\= x = 173.2$

Generally the standard deviation is mathematically represented as

$\sigma = \sqrt{\frac{\sum (x_i - \= x )^2 }{n} }$

=> $\sigma = \sqrt{\frac{ (156 - 173.2 )^2 + (136 - 173.2 )^2 + \cdots + (181- 173.2 )^2 }{25} }$

=> $\sigma = 24.261$

Generally the test statistics is mathematically represented as

$t = \frac{\= x - \mu }{\frac{\sigma}{\sqrt{n} } }$

=> $t = \frac{173.2 - 180 }{\frac{24.261}{\sqrt{25} } }$

=> $z = -1.40$

Generally p-value is mathematically represented as

$p-value = P(Z < z)$

=> $p-value = P(Z < -1.40)$

From the z-table

$P(Z < -1.40) = 0.080757$

=> $p-value = 0.080757$

From the obtained value we that $p-value > \alpha$

The decision rule is fail to reject the null hypothesis

The conclusion

There no sufficient evidence to support the vendor claim that the warehouse management system reduces the average pick, pack, and ship time to below 3 minutes(180 seconds) per order through bin location and routing optimization

5 0

3 years ago

I need answers for A -F

mestny [16]

Answer:

A. 103º

B. 96º

C. m<4 = m<10

D. m<1 = m<16

E. m<5 = m<13

F. m<14 + m<15 = 180º

(Not entirely sure for C-F... But I hope this helps!)

8 0

3 years ago

What inequality describes the weight range of a single nickel

Evgen [1.6K]

Answer:

4.806 g ≤ n ≤ 5.194 where n is the mass of a nickle.

Step-by-step explanation:

The problem tells us the weight can vary by .194 g, this means it can be up to .194 lighter and 194 heavier. Well what value is .194 lighter and .194 heavier? 4.806 and 5.194 respectively. So that means a nickle can be anywhere in that range, which tells us the inequality.

5 0

3 years ago

A lawn mower cost $470 when new and is expected to last 10 years. What will it be worth in 7.5 years? (Assume the lawn mower has

USPshnik [31]

Answer: 117.5
Explanation: Assuming that after 10 years the lawn mower would have a value of 0. We can evenly divide $470 by the amount of years.
$750/10 = $75
We would then multiply 75*7.5 years to then realize the value lost over that amount of time would be $352.5, subtracting that from the original value of $470 would then be 117.5

5 0

3 years ago