Question

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) that......a single Rayovac battery lasts under 22 hours: ...the mean "continuous use-life" of 4 randomly selected Rayovac batteries is less than 22 hours: ...the mean "continuous use-life" of 16 randomly selected Rayovac batteries is less than 22 hours: ...the mean "continuous use-life" of 64 randomly selected Rayovac batteries is less than 22 hours: ...a single Duracell battery lasts under 22 hours: ...the mean "continuous use-life" of 4 randomly selected Duracell batteries is less than 22 hours: ...the mean "continuous use-life" of 9 randomly selected Duracell batteries is less than 22 hours: ...the mean "continuous use-life" of 25 randomly selected Duracell batteries is less than 22 hours: ...a randomly selected Rayovac battery lasts longer than a randomly selected Duracell battrey: ...given random samples of four of each kind, the Rayovac batteries have a longer mean "continuous use-life" than do the Duracell batteries:

Answer 1

Answer:

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.

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