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Feliz [49]
3 years ago
10

5 squirrels were found to have an average weight of 8.9 ounces with a sample standard deviation is 0.9. Find the 95% confidence

interval of the true mean weight.
Mathematics
1 answer:
stira [4]3 years ago
7 0

Answer: ( 8.11 , 9.69)

the 95% confidence interval of the true mean weight

= ( 8.11 , 9.69)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 8.9

Standard deviation r = 0.9

Number of samples n = 5

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

8.9+/-1.96(0.9/√5)

8.9+/-1.96(0.4025)

8.9+/- 0.79

= ( 8.11 , 9.69)

Therefore at 95% confidence interval (a,b) = ( 8.11 , 9.69)

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Answer:

P(52

And we can find this probability with this difference:

P(-0.851

And if we use the normal standard distribution or excel we got:

P(-0.851

Step-by-step explanation:

Let X the random variable that represent the time required to construct and test a particular component of a population, and for this case we know the distribution for X is given by:

X \sim N(60,9.4)  

Where \mu=60 and \sigma=9.4

We want to find this probability:

P(52

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

Using this formula we got:

P(52

And we can find this probability with this difference:

P(-0.851

And if we use the normal standard distribution or excel we got:

P(-0.851

5 0
3 years ago
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