PLEASE HELP What is the rule for the sequence with the first four terms below?
1 answer:
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Answer:
Probability that the proportion in our sample of red candies will be less than 20% is 0.5 .
Step-by-step explanation:
We are given that 20% of the candy produced are red. A random sample of 100 bags of Skittles is taken.
The distribution we can use here is;
~ N(0,1)
where, p = 0.20 and n = 100
Let = proportion of red candies in our sample
So, P( < 0.20) = P(
<
) = P(Z < 0) = 0.5
Therefore, probability that the proportion in our sample of red candies will be less than 20% is 0.5 .
Answer:
Variance = 4.68
Step-by-step explanation:
The formula for the variance is:
Where:
The mean can be calculated as each value multiplied by its probability
Replacing the mean and the summatory of X:
Answer:
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 = [ ( × σ ) / E ]²
given that confidence level = 95%, so = 1.96
Now,
a) Margin of error E = 500
n = [ ( × σ ) / E ]²
n = [ ( 1.96 × 10000 ) / 500 ]²
n = [ 19600 / 500 ]²
n = 1536.64 ≈ 1537
b) Margin of error E = 200
n = [ ( × σ ) / E ]²
n = [ ( 1.96 × 10000 ) / 200 ]²
n = [ 19600 / 200 ]²
n = 9604
c) Margin of error E = 100
n = [ ( × 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.