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4 years ago

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A manufacturing process outputs parts having a normal distribution with a mean of 30 cm and standard deviation of 2 cm. From a p

roduction sample of 80 parts, what proportion of the sample can be expected to fall between 28 and 32 cm

1 answer:

AfilCa [17]4 years ago

7 0

Answer:

68% of the sample can be expected to fall between 28 and 32 cm

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 30

Standard deviation = 2

What proportion of the sample can be expected to fall between 28 and 32 cm

28 = 30 - 2

28 is one standard deviation below the mean

32 = 30 + 2

32 is one standard deviation above the mean.

By the Empirical Rule, 68% of the sample can be expected to fall between 28 and 32 cm

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One more question on this test guys. please answer as fast as possible. If you do,

GaryK [48]

1. The given rectangular equation is $x=2$ .

We substitute $x=r\cos \theta$ .

$r\cos \theta=2$

Divide through by $\cos \theta$

$r=\frac{2}{\cos \theta}$

$r=2}\sec \theta$

$\boxed{x=2\to r=2\sec \theta}$

2. The given rectangular equation is:

$x^2+y^2=36$

This is the same as:

$x^2+y^2=6^2$

We use the relation $r^2=x^2+y^2$

This implies that:

$r^2=6^2$

$\therefore r=6$

$\boxed{x^2+y^2=36\to r=6}$

3. The given rectangular equation is:

$x^2+y^2=2y$

This is the same as:

We use the relation $r^2=x^2+y^2$ and $y=r\sin \theta$

This implies that:

$r^2=2r\sin \theta$

Divide through by r

$r=2\sin \theta$

$\boxed{x^2+y^2=2y\to r=2\sin \theta}$

4. We have $x=\sqrt{3}y$

We substitute $y=r\sin \theta$ and $x=r\cos \theta$

$r\cos \theta=r\sin \theta\sqrt{3}$

This implies that;

$\tan \theta=\frac{\sqrt{3}}{3}$

$\theta=\frac{\pi}{6}$

$\boxed{x=\sqrt{3}y\to \theta=\frac{\pi}{6}}$

5. We have $x=y$

We substitute $y=r\sin \theta$ and $x=r\cos \theta$

$r\cos \theta=r\sin \theta$

This implies that;

$\tan \theta=1$

$\theta=\frac{\pi}{4}$

$\boxed{x=y\to \theta=\frac{\pi}{4}}$

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3 years ago

A small bag of flavor weighed 13 ounces. A large bag was 5 percent heavier.How much does the large bag weigh?

Novay_Z [31]

13+5=18
The large bag would be 18 because you have to add 5 more ounces than 13

5 0

4 years ago

A simple random sample of 50 items from a population with 7 resulted in a sample mean of 35.

Vilka [71]

The 90% , 99% confidence interval for the population mean is 32.145 < $\rm \mu$ < 35.855 and 31.093 < $\rm \mu$ < 36.907

<h3>What is Probability ?</h3>

Probability is the study of likeliness of an event to happen.

It is given that

Total Population = 50

Mean = 35

The confidence interval is given by

$\rm CI = \mu + z \dfrac{s}{\sqrt n}$

$\rm \mu$ is the mean

z is the confidence level value

s is the standard deviation

n is the population width

(a) The 90% confidence interval for the population mean

90% $\rm \alpha / 2$ = 0.05

Z = 1.64

34 $\rm \pm$ 1.64 * 8 / √50

34 $\rm \pm$ 1.855

32.145 < $\rm \mu$ < 35.855

(b) The 99% confidence interval for the population mean

99% $\rm \alpha / 2$ = 0.005

Z=2.57

34 $\rm \pm$ 2.57 * 8 / √50

34 $\rm \pm$ 2.907

31.093 < $\rm \mu$ < 36.907

Therefore the confidence interval for population mean has been determined.

The complete question is

A simple random sample of 50 items from a population width =7 resulted in a sample mean of 35. If required, round your answers to two decimal places.

a. Provide a 90% confidence interval for the population mean

b. Provide a 99% confidence interval for the population mean

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