Mathematics, 28.03.2020 04:01 peytonbrien2002

The following sample of lengths was taken from 9 rods off the assembly line. Construct the 99% confidence interval for the population standard deviation for all rods that come off the assembly line. Round your answers to two decimal places.

13.6,13.8,14.1,13.6,13.3,13.5,13.9, 13.3,14.1

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Answer from: Blossom824

99% confidence interval for the population standard deviation = (0.17 , 0.75).

Step-by-step explanation:

We are given that the following sample of lengths was taken from 9 rods off the assembly line;

13.6, 13.8, 14.1, 13.6, 13.3, 13.5, 13.9, 13.3, 14.1

So, firstly the pivotal quantity for 99% confidence interval for the population standard deviation is given by;

P.Q. = $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, s = sample standard deviation

$\sigma$ = population standard deviation

n = sample of rods = 9

Also, $s^{2} = \frac{\sum (X-\bar X)^{2} }{n-1}$ , where X = individual data value

$\bar X$ = mean of data values = 13.7

$s^{2}$ = 0.094

So, 99% confidence interval for population standard deviation, is;

P(1.344 < $\chi^{2} __8$ < 21.95) = 0.99 {As the table of $\chi^{2}$ at 8 degree of freedom

gives critical values of 1.344 & 21.95}

P(1.344 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 21.95) = 0.99

P( $\frac{ 1.344}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{ 21.95}{(n-1)s^{2} }$ ) = 0.99

P( $\frac{ (n-1)s^{2}}{21.95 }$ < $\sigma^{2}$ < $\frac{ (n-1)s^{2}}{1.344 }$ ) = 0.99

99% confidence interval for $\sigma^{2}$ = ( $\frac{ (n-1)s^{2}}{21.95 }$ , $\frac{ (n-1)s^{2}}{1.344 }$ )

= ( $\frac{ (9-1)\times 0.094}{21.95 }$ , $\frac{ (9-1)\times 0.094}{1.344 }$ )

= (0.03 , 0.56)

99% confidence interval for $\sigma$ = ( $\sqrt{0.03}$ , $\sqrt{0.56}$ )

= (0.17 , 0.75)

Therefore, 99% confidence interval for the population standard deviation for all rods that come off the assembly line is (0.17 , 0.75).

Answer from: Quest

geometric series converge the common ratio r has magnitude less than one.

$|r| < 1$

which is equivalent to

$-1 < r < 1$

for the example we need

|2-x| < 1

-1 < 2-x and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1 diverges

1b. r=50/100=1/2 converges

when these converge they converge to

$s = \dfrac{a}{1 -r}$

where a is the first term.

$s = \dfrac{100}{1 - 1/2} = 200$

answer: 200

1c. r=-1 so it's not the case that |r| < 1. diverges

2a. x is the common ratio so we converge when |x|< 1

2b. r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c. r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

$|x| < \frac 1 2 \sqrt[3]{2}$

Answer from: Quest

ghjmutmum

step-by-step explanation:

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