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amid [387]
3 years ago
11

A study studied the birth weights of 1,278 babies born in the united states. the mean weight was 3234 grams with a standard devi

ation of 871 grams. assume that birth weight data are approximately bell-shaped. estimate the number of newborns who weighed between 1492 grams and 4976 grams.
Mathematics
2 answers:
babymother [125]3 years ago
6 0
<span>1220 Subtracting the lower boundary of 1492 grams from the mean of 3234 gives you 1742 grams below the mean. Dividing 1742 by the standard deviation of 871 gives you 2 standard deviations below the curve. Now doing the same with the upper limit of 4976 grams also gives you 2 standard deviations above the mean (4976-3234)/871 = 2 So you now look for what percentage of the population lies within 2 standard deviations of the mean. Standard lookup tables will indicate that 95.4499736% of the population will be within 2Ď of the mean. So multiply 1278 by 0.954499736 giving 1219.851. Then round to the nearest whole number and you have an estimated 1220 babies that weigh between 1492 grams and 4976 grams.</span>
hram777 [196]3 years ago
6 0

The number of newborns who weighed between 1492 grams and 4976 grams are \boxed{1219}.

Further Explanation:

Given:

The number of babies born in the United States is 1278.

The mean weight was 3234{\text{ gram}}.

The standard deviation is 871{\text{ gram}}.

Explanation:

The empirical rule is used to check that the given data set is normally distributed. The data values should lie within the three standard deviation of the mean.

Empirical rule is defined as follows,

68% data points will lie within the first standard deviation of the mean.

\boxed{P\left( {\mu  - \sigma } \right) = 68\% }

95.4% data points will lie within the two standard deviation of the mean.

\boxed{P\left( {\mu  - 2\sigma } \right) = 95.4\% }

99.7% data points will lie within the three standard deviation of the mean.

\boxed{P\left( {\mu  - 3\sigma }\right) = 99.7\% }

Here, \mu represents the mean and \sigma represents the standard deviation.

Subtract or add two times the standard deviation from the mean then the interval can be expressed as,

\begin{aligned}{\text{Interval}} &= \mu\pm 2\sigma\\&= 3234 \pm 2 \times 871\\&= 3234 \pm 1742\\&= \left( {1492,4976} \right)\\\end{aligned}

The number of newborns who weighed between 1492 grams and 4976 grams are \boxed{1219}.

Learn more:

  1. Learn more about normal distribution brainly.com/question/12698949
  2. Learn more about standard normal distribution brainly.com/question/13006989
  3. Learn more about confidence interval of meanhttps://brainly.com/question/12986589

Answer details:

Grade: College

Subject: Statistics

Chapter: Confidence Interval

Keywords: study, studied, birth weights, babies,  mean weight, standard deviation, bell-shaped, mean, repeating, indicated, normal distribution, percentile, percentage, undesirable behavior, proportion, empirical rule.

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Harlamova29_29 [7]

Answer:

t=\frac{47-50}{\frac{5.5}{\sqrt{30}}}=-2.988    

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If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is significantly different from 50 mpg at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=47 represent the sample  mean  

s=5.5 represent the sample standard deviation

n=30 sample size  

\mu_o =50 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is 50 or no, the system of hypothesis would be:  

Null hypothesis:\mu = 50  

Alternative hypothesis:\mu \neq 50  

If we analyze the size for the sample is = 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{47-50}{\frac{5.5}{\sqrt{30}}}=-2.988    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=30-1=29  

Since is a two tailed test the p value would be:  

p_v =2*P(t_{(29)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is significantly different from 50 mpg at 5% of signficance.  

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If it takes you 40 minutes to go 20 miles downstream and then 60 minutes on the way back, what is the speed of the current?
IrinaK [193]

Answer:

Speed of Current = 5 miles per hour

Step-by-step explanation:

We know distance formula,

D = RT

Where

D is distance

R is rate

T is time

If we let speed of boat (assume) to be "x" and speed of current to be "c"

Then downstream rate is (with current) = x + c

Upstream rate is (against current) = x - c

40 mins to go 20 miles downstream, that means:

D = RT

20 = (x + c)(40)

and

60 minutes to go upstream, 20 miles, that means:

D = RT

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Simplifying first equation:

40x + 40c = 20

Simplifying second equation:

60x - 60c = 20

Multiplying first equation by 60, we get:

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Multiplying second equation by 40, we get:

40 * [60x - 60c = 20] = 2400x - 2400c = 800

Now we add up both these equations:

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Which is not a correct way to rewrite this expression using the distributive
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Given:

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Using distributive property, we get

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Using distributive property the given expression can rewritten as:

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Only the expression in option A is not a correct way to rewrite the given expression because (2x^2+4x-7) is not distributed to (x-2) properly.

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