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Roman55 [17]
2 years ago
9

We want to estimate the population mean within 5, with a 99% level of confidence. the population standard deviation is estimated

to be 15. how large a sample is required?
Mathematics
1 answer:
pashok25 [27]2 years ago
3 0
A sample size of 60 is required.

We use the formula
n=(\frac{z*\sigma}{E})^2

We first find the z-score associated with this level of confidence:
Convert 99% to a decimal:  99/100 = 0.99
Subtract from 1:  1-0.99 = 0.01
Divide by 2:  0.01/2 = 0.005
Subtract from 1:  1-0.005 = 0.995

Using a z-table (http://www.z-table.com) we see that this value is equally distant from 2.57 and 2.58; therefore we will use 2.575:

n=(\frac{15(2.575)}{5})^2=59.68\approx60
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A machine fills boxes weighing Y lb with X lb of salt, where X and Y are normal with mean 100 lb and 5 lb and standard deviation
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Answer:

Option b. None is the correct option.

The Answer is 63%

Step-by-step explanation:

To solve for this question, we would be using the z score formula

The formula for calculating a z-score is given as:

z = (x-μ)/σ,

where

x is the raw score

μ is the population mean

σ is the population standard deviation.

We have boxes X and Y. So we will be combining both boxes

Mean of X = 100 lb

Mean of Y = 5 lb

Total mean = 100 + 5 = 105lb

Standard deviation for X = 1 lb

Standard deviation for Y = 0.5 lb

Remember Variance = Standard deviation ²

Variance for X = 1lb² = 1

Variance for Y = 0.5² = 0.25

Total variance = 1 + 0.25 = 1.25

Total standard deviation = √Total variance

= √1.25

Solving our question, we were asked to find the percent of filled boxes weighing between 104 lb and 106 lb are to be expected. Hence,

For 104lb

z = (x-μ)/σ,

z = 104 - 105 / √25

z = -0.89443

Using z score table ,

P( x = z)

P ( x = 104) = P( z = -0.89443) = 0.18555

For 1061b

z = (x-μ)/σ,

z = 106 - 105 / √25

z = 0.89443

Using z score table ,

P( x = z)

P ( x = 106) = P( z = 0.89443) = 0.81445

P(104 ≤ Z ≤ 106) = 0.81445 - 0.18555

= 0.6289

Converting to percentage, we have :

0.6289 × 100 = 62.89%

Approximately = 63 %

Therefore, the percent of filled boxes weighing between 104 lb and 106 lb that are to be expected is 63%

Since there is no 63% in the option, the correct answer is Option b. None.

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Ben swims​ 50,000 yards per week in his practices. Given this amount of​ training, he will swim the​ 100-yard butterfly in 51.5
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Answer:

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Then we can interpret this as he will reduce his time an <em>additional </em>0.0002 seconds for every <em>additional </em>yard he trains.

This is an approximation that is valid in the interval of 60,000 to  70,000 yards of training.

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