Mathematics, 10.11.2019 04:31 ghernadez

Rayovac type-d batteries have a "continuous use-life" that is approximately normally distributed with mean 21 hours and standard deviation 1.38 hours. duracell type-d batteries have "continuous use-life" that is approximately normally distributed with mean 24 hours and standard deviation 2.13 hours. find the probability (to four places after the decimal) single rayovac battery lasts under 22 hours: mean "continuous use-life" of 4 randomly selected rayovac batteries is less than 22 hours: mean "continuous use-life" of 16 randomly selected rayovac batteries is less than 22 hours: mean "continuous use-life" of 64 randomly selected rayovac batteries is less than 22 hours: single duracell battery lasts under 22 hours: mean "continuous use-life" of 4 randomly selected duracell batteries is less than 22 hours: mean "continuous use-life" of 9 randomly selected duracell batteries is less than 22 hours: mean "continuous use-life" of 25 randomly selected duracell batteries is less than 22 hours: randomly selected rayovac battery lasts longer than a randomly selected duracell battrey: random samples of four of each kind, the rayovac batteries have a longer mean "continuous use-life" than do the duracell batteries:

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Answer from: moneybaggtjuan

See below

Step-by-step explanation:

Rayovac type-D batteries

Mean

$\bf \bar x_1$ = 21 h

Standard deviation

$\bf s_1$ = 1.38 h

Find the probability (to four places after the decimal) that a single Rayovac battery lasts under 22 hours:

Here we use the Normal N(0,1)

$\bf P(z$

To find this value use either a table or a spreadsheet.

In Excel

NORMDIST(0.7246,0,1,1)

In OpenOffice Calc

NORMDIST(0.7246;0;1;1)

Whenever we have a sample size less than 30, we will be using the Student's t distribution with (sample size -1) degrees of freedom instead of the Normal.

Find the probability that the mean "continuous use-life" of 4 randomly selected Rayovac batteries is less than 22 hours:

$\bf P(t$

To find this value use either a table or a spreadsheet.

In Excel

1-TDIST(1.4493,3,1)

In OpenOffice Calc

1-TDIST(1.4493;3;1)

Find the probability that the mean "continuous use-life" of 16 randomly selected Rayovac batteries is less than 22 hours:

$\bf P(t$

To find this value use either a table or a spreadsheet.

In Excel

1-TDIST(2.8986,15,1)

In OpenOffice Calc

1-TDIST(2.8986 ;15;1)

Find the probability that the mean "continuous use-life" of 64 randomly selected Rayovac batteries is less than 22 hours:

Since the sample size is greater than 30 we can use back the Normal

$\bf P(z$

To find this value use either a table or a spreadsheet.

In Excel

NORMDIST(5.7971,0,1,1)

In OpenOffice Calc

NORMDIST(5.7971;0;1;1)

Duracell type-D batteries

Mean

$\bf \bar x_2$ = 24 h

Standard deviation

$\bf s_2$ = 2.13 h

Find the probability that a single Duracell battery lasts under 22 hours

Here we use the Normal N(0,1)

$\bf P(z$

To find this value use either a table or a spreadsheet.

In Excel

NORMDIST(-0.9390,0,1,1)

In OpenOffice Calc

NORMDIST(-0.9390;0;1;1)

Find the probability that the mean "continuous use-life" of 4 randomly selected Duracell batteries is less than 22 hours:

$\bf P(t$

To find this value use either a table or a spreadsheet.

In Excel

TDIST(1.8779,3,1)

In OpenOffice Calc

TDIST(1.8779;3;1)

Find the probability that the mean "continuous use-life" of 9 randomly selected Duracell batteries is less than 22 hours:

$\bf P(t$

To find this value use either a table or a spreadsheet.

In Excel

TDIST(2.8169,8,1)

In OpenOffice Calc

TDIST(2.8169;8;1)

Find the probability that the mean "continuous use-life" of 25 randomly selected Duracell batteries is less than 22 hours:

$\bf P(t$

To find this value use either a table or a spreadsheet.

In Excel

TDIST(4.6948,24,1)

In OpenOffice Calc

TDIST(4.6948;24;1)

Find the probability that a randomly selected Rayovac battery lasts longer than a randomly selected Duracell battery given random samples of four of each kind.

$\bf P(t\frac{\bar x_1-\bar x_2}{\sqrt{s_1^2/4+s_2^2/4}})=P(t\frac{21-24}{\sqrt{(1.38)^2/4+(2.13)^2/4}})=\\=P(t-2.364)=1-P(t$

Answer from: Quest

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Answer from: Quest

The answer would be a

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