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leonid [27]
3 years ago
13

A smoothie shop has 40 stores and 55% of the stores are in california the rest of the stores are in nevada.how many are in nevad

a?
Mathematics
1 answer:
blondinia [14]3 years ago
4 0
Given:
40 stores in total
55% of the stores are in California, the rest are in Nevada.

This is simply multiplying the corresponding percentage from the whole to get its number and deducting the number to get the required number.

40 represents 100%.

40 x 55% = 22 stores in California
40 - 22 = 18 stores in Nevada. 

There are 18 stores in Nevada.

Since 55% represents  California, then, 45% represents Nevada.

40 * (100%  - 55%) = 40 * 45% = 18 stores.
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1) the planning value for the population standard deviation is 10,000

2)

a) Margin of error E = 500, n = 1536.64 ≈ 1537

b) Margin of error E = 200, n = 9604

c) Margin of error E = 100, n = 38416

3)

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

Step-by-step explanation:

Given the data in the question;

1) Planning Value for the population standard deviation will be;

⇒ ( 50,000 - 10,000 ) / 4

= 40,000 / 4

σ = 10,000

Hence, the planning value for the population standard deviation is 10,000

2) how large a sample should be taken if the desired margin of error is;

we know that, n = [ (z_{\alpha /2 × σ ) / E ]²

given that confidence level = 95%, so z_{\alpha /2  = 1.96

Now,

a) Margin of error E = 500

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 500 ]²

n = [ 19600 / 500 ]²

n = 1536.64 ≈ 1537

b) Margin of error E = 200

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 200 ]²

n = [ 19600 / 200 ]²

n = 9604

c)  Margin of error E = 100

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 100 ]²

n = [ 19600 / 100 ]²

n = 38416

3) Would you recommend trying to obtain the $100 margin of error?

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

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