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3 years ago

8

kelly subscribes to a service called mysongs for downloading music she pays a $7 fee each month to download an unlimited numbers

of songs at $0.20 each last month kelly's bill was $23.80

1 answer:

Nastasia [14]3 years ago

8 0

subtract the fee from the total to get the amount actually spent on songs

23.80-7 = 16.80

divide that amount by price per song to get number of songs downloaded

16.80 /0.20 = 84 songs

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Assuming that the population is normally distributed, construct a 9090% confidence interval for the population mean for each o

jasenka [17]

Given:

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Standard dev: 1.9

Set B:

Mean: 4.5

Standard dev: 2.45

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Set A:

Standard Error, SE = s/ √n = 1.9/√8 = 0.67

Degrees of freedom = n - 1 = 8 -1 = 7

t- score = 1.89457861

<span> <span><span> <span> </span> </span> </span></span>

Width of the confidence interval = t * SE = 1.89457861* 0.67 = 1.272685913

Lower Limit of the confidence interval = x-bar - width = 4.5 - 1.272685913 = 3.23

Upper Limit of the confidence interval = x-bar + width = 4.5 + 1.272685913 = 5.77

The 90% confidence interval is [3.23, 5.77]

Set B:

Standard Error, SE = s/ √n = 2.45/√8 = 0.87

Degrees of freedom = n - 1 = 8 -1 = 7

t- Score = 1.89457861

<span> <span><span> <span> </span> </span> </span></span>

Width of the confidence interval = t * SE = 1.89457861* 0.87 = 1.641094994

Lower Limit of the confidence interval = x-bar - width = 4.5 - 1.641094994 = 2.86

Upper Limit of the confidence interval = x-bar + width = 4.5 + 1.641094994 = 6.14

The 90% confidence interval is [2.86, 6.14]

<span>We can obviously see that sample B has more variation in the scores than sample A. The fact that the standard deviation is 2.45 for B and 1.9 for A). Therefore, they yield dissimilar confidence intervals even though they have the same mean and range.</span>

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X^2-22x-48=0
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