Write five expressions equivalent to 20x + 100

A sample of 18 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is n

ki77a [65]

Answer:

a)

i. x=- - - - - -

ii. o=- - - - - -

iii. Sx=- - - - -

b) The random variable X signifies the weights of the bags of candies which are selected at random.

c) The statistics variable X is a measure on a sample that is used as an estimate of the population mean.X is the mean weight of the 16 bags of candies that were selected.

d) The distribution needed for this problem is normal distribution with parameters because the population standard deviation is known.

N(X, o/ $\sqrt{n}$ )

So, the distribution is

N(2, 0.12/ $\sqrt{16}$ )

e)

i. The 90% confidence interval for the population mean weight of the candies is 1.9589 , 2.0411

iii. The error bound is 0.0411

f)

i. The 98% confidence interval for the population mean weight of the candies is 1.9418 , 2.0582

iii. The error bound is 0.0582

g) The difference in the confidence intervals in part (f) and part (e) is of the level of confidence. The change is, the change in the area being calculated for the normal distribution. Therefore, the larger confidence level results in larger area and larger interval.

Hence the interval in part is larger than the one in part (e).

h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.

Step-by-step explanation:

a)

i. x=- - - - - -

ii. o=- - - - - -

iii. Sx=- - - - -

b) The random variable X signifies the weights of the bags of candies which are selected at random.

c) The statistics variable X is a measure on a sample that is used as an estimate of the population mean.X is the mean weight of the 16 bags of candies that were selected.

d) The distribution needed for this problem is normal distribution with parameters because the population standard deviation is known.

N(X, o/ $\sqrt{n}$ )

So, the distribution is

N(2, 0.12/ $\sqrt{16}$ )

e)

i. The 90% confidence interval for the population mean weight of the candies is 1.9589 , 2.0411

iii. The error bound is 0.0411

f)

i. The 98% confidence interval for the population mean weight of the candies is 1.9418 , 2.0582

iii. The error bound is 0.0582

g) The difference in the confidence intervals in part (f) and part (e) is of the level of confidence. The change is, the change in the area being calculated for the normal distribution. Therefore, the larger confidence level results in larger area and larger interval.

Hence the interval in part is larger than the one in part (e).

h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.

5 0

3 years ago

Jean wants to put furniture in her clubhouse. She drew a floor plan of the clubhouse, as shown. Each grid unit represents one fo

Softa [21]

I think its A) Which polygon names the shape of the floor?

3 0

4 years ago

No thanks i got it i was just joking

Anton [14]

Step-by-step explanation:

why would you joke I was willing to help you hun but enjoy the rest of your day

5 0

3 years ago

3x + 1 − 4x3 + 6x6 −2x2 standard form

Molodets [167]

7 0

3 years ago

A manufacturer produces piston rings for an automobile engine. It is known that ring diameter is normally distributed with σ=0.0

oksian1 [2.3K]

<h2>Answer with explanation:</h2>

The confidence interval for population mean is given by :-

$\overline{x}\pm z^* SE$ (1)

, where $\overline{x}$ = sample mean

z* = critical value.

SE = standard error

and $SE=\dfrac{\sigma}{\sqrt{n}}$ , $\sigma$ = population standard deviation.

n= sample size.

As per given , we have

$\overline{x}=74.021$

$\sigma=0.001$

n= 15

It is known that ring diameter is normally distributed.

$SE=\dfrac{0.001}{\sqrt{15}}=0.000258198889747\approx0.000258199$

By z-table ,

The critical value for 95% confidence = z*= 1.96

A 99% two-sided confidence interval on the true mean piston diameter :

$74.021\pm (2.576) (0.000258199)$ (using (1))

$74.021\pm 0.000665120624$

$74.021\pm 0.000665120624\\\\=(74.021- 0.000665120624,\ 74.021+ 0.000665120624)\\\\=(74.0203348794,\ 74.0216651206)\approx(74.020,\ 74.022)$ [Rounded to three decimal places]

∴ A 99% two-sided confidence interval on the true mean piston diameter = (74.020, 74.022)

By z-table ,

The critical value for 95% confidence = z*= 1.96

A 95% lower confidence bound on the true mean piston diameter:

$74.021- (1.96) (0.000258199)$ (using (1))

$74.021- 0.00050607004=74.02049393\approx74.020$ [Rounded to three decimal places]

∴ A 95% lower confidence bound on the true mean piston diameter= 74.020

7 0

3 years ago