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yulyashka [42]
3 years ago
13

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of

65 weekly reports showed a sample mean of 17.5 customer contacts per week. The sample standard deviation was 5.7. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel. 90% Confidence interval, to 2 decimals: _______
Mathematics
1 answer:
Aleks [24]3 years ago
3 0

Answer:

CI for 90% = ( 16.34, 18.66)

Therefore at 90% confidence interval (a,b) = ( 16.34, 18.66)

And,

CI for 95% = ( 16.11, 18.89)

Therefore at 95% confidence interval (a,b) = ( 16.11, 18.89)

Step-by-step explanation:

Answer: = ( 2.64, 3.14)

Therefore at 95% confidence interval (a,b) = ( 2.64, 3.14)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 17.5

Standard deviation r = 5.7

Number of samples n = 65

Confidence interval = 90% and 95%

z(at 90% confidence) = 1.645

z(at 95% confidence) = 1.96

Substituting the values we have; for 90%

17.5+/-1.645(5.7/√65)

17.5+/-1.645(0.707)

17.5 +/- 1.16

= ( 16.34, 18.66)

Therefore at 90% confidence interval (a,b) = ( 16.34, 18.66)

Substituting the values we have; for 95%

17.5+/-1.96(5.7/√65)

17.5+/-1.96(0.707)

17.5 +/- 1.39

= ( 16.11, 18.89)

Therefore at 95% confidence interval (a,b) = ( 16.11, 18.89)

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Answer:

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3 years ago
A college conducts a common test for all the students. For the Mathematics portion of this test, the scores are normally distrib
Jet001 [13]

Using the normal distribution, it is found that 58.97% of students would be expected to score between 400 and 590.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

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

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

\mu = 502, \sigma = 115

The proportion of students between 400 and 590 is the <u>p-value of Z when X = 590 subtracted by the p-value of Z when X = 400</u>, hence:

X = 590:

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

Z = \frac{590 - 502}{115}

Z = 0.76

Z = 0.76 has a p-value of 0.7764.

X = 400:

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

Z = \frac{400 - 502}{115}

Z = -0.89

Z = -0.89 has a p-value of 0.1867.

0.7764 - 0.1867 = 0.5897 = 58.97%.

58.97% of students would be expected to score between 400 and 590.

More can be learned about the normal distribution at brainly.com/question/27643290

#SPJ1

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Answer:

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Step-by-step explanation:

The probability of at least one success is

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where,

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'n' is number of events

We have probability of getting 2 is 1/6 thus 'p' = 1/6

Applying values we get

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