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inysia [295]
3 years ago
12

To test upper h 0 : mu equals 72 versus upper h 1 : mu less than 72​, a simple random sample of size nequals20 is obtained from

a population that is known to be normally​ distributed, and the sample standard deviation is found to be 6. complete parts ​(a) and ​(b) below.
Mathematics
1 answer:
katen-ka-za [31]3 years ago
6 0
We solve using equation \beta   \sqrt{x}  \sqrt[n]{x}  \lim_{n \to \infty} a_n  \alpha⇔ ∈ ω. ㏒
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Find the probability of rolling a prime number when a 16-sided die is rolled.
Katen [24]

Answer:

6/16

Step-by-step explanation:

On a 16 sided die there would be number as follows :

1,2,3,.. 16 The prime number out in this set of numbers are

2,3,5,7,11, and 13

There are 6 prime number so the probability of rolling a prime number is

6/16

4 0
3 years ago
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27 millimeters multiplied by 1 cm/10 mm
Basile [38]
27 millimeters multiplied by 1 cm/10 mm= 2.7 centimeters

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3 years ago
Find the indicated probability or percentage for the sampling error. The distribution of weekly salaries at a large company is r
Flauer [41]

Answer:

The probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 80 employees will be at most $75 is 0.9297.

Step-by-step explanation:

According to the Central Limit Theorem if we have a non-normal population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample means is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

\mu=\$1000\\\sigma=\$370\\n=80

As <em>n</em> = 80 > 30, the central limit theorem can be used to approximate the sampling distribution of sample mean weekly salaries.

Let \bar X represent the sample mean weekly salaries.

The distribution of \bar X is: \bar X\sim N(\$1000,\ \$41.37)

Now we need to compute the probability of the sampling error made in estimating the mean weekly salary to be at most $75.

The sampling error is the the difference between the estimated value of the parameter and the actual value of the parameter, i.e. in this case the sampling error is, |\bar X-\mu|= 75.

Compute the probability as follows:

P(-75

                                     =P(-1.81

Thus, the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 80 employees will be at most $75 is 0.9297.

3 0
3 years ago
A small train at an amusement park is carrying passengers. Half of the passengers are 4 years old, 1/10 are 5 years old, 3/10 ar
drek231 [11]
Your answer is 5, because the majority is 5 years old

8 0
3 years ago
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I need help with this please​
harina [27]

Answer:

acute

Step-by-step explanation:q

7 0
3 years ago
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