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Liono4ka [1.6K]
3 years ago
8

An SRS of 18 recent birth records at the local hospital was selected. In the sample, the average birth weight was 119.6 ounces a

nd the standard deviation was 6.5 ounces. Assume that the population of birth weights of all babies born in this hospital follow a Normal distribution, with mean μ.
The standard error of the mean is
A. 6.50.
B. 1.53.
C. 0.36.
D. 0.02.
Mathematics
1 answer:
Crazy boy [7]3 years ago
8 0
The standard error of the mean is the standard deviation which is 6.50
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-5x-6(-6+3x)=105 what is the answer
lyudmila [28]

Answer:

x = -3

Step-by-step explanation:

expand -23x + 36 = 105

subtract 36 from both sides -23x +36 -36 = 105 - 36

Simplify -23x = 69

Divid both sides by -23: -23x / - 23 = 69 / -23

x = -3

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What is 3x+5Y=15for y
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3 years ago
Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,
valentinak56 [21]

Answer:

The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 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 mean is given by,

\mu_{\bar x}=\mu

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

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

The information provided is:

<em>μ</em> = 144 mm

<em>σ</em> = 7 mm

<em>n</em> = 50.

Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.

\bar X\sim N(\mu_{\bar x}=144, \sigma_{\bar x}^{2}=0.98)

Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:

P(\bar X-\mu_{\bar x}>2.6)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}} >\frac{2.6}{\sqrt{0.98}})

                           =P(Z>2.63)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

8 0
2 years ago
2. It costs Edgar $2.65 one way to ride the bus to work. He uses the bus to go to and from work four days per week. He buys a bo
Maurinko [17]
$27

2.65 x 2 (since he has to go both to and from work each day) = $5.30
Add the daily cost of the water bottle (1.45) to make his total cost for one day, which is 6.75
Multiply this number by 4 (because he goes 4 days a week) to get $27 for one week.
6 0
2 years ago
What is 1.65 as a fraction
marshall27 [118]

33/20  as well as 1 and 13/20. First you look it up in the internet aka Google.

4 0
2 years ago
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