Answer:
The standard deviation would have to be 0.05.
Step-by-step explanation:
We can solve this problem using the 68-95-99.7 rule for normal distributions:
The rule states that:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviations of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
What would the value of σ have to be to ensure that 95% of all readings are within 0.1° of μ?
This means that 2 standard deviations would need to be within 0.1 of the mean. So
The standard deviation would have to be 0.05.
A=4-3b
subtract 4 from both sides
a-4=-3b
divide by -3 for both sides
(a-4)/-3 = b
If inspection department wants to estimate the mean amount with 95% confidence level with standard deviation 0.05 then it needed a sample size of 97.
Given 95% confidence level, standard deviation=0.05.
We know that margin of error is the range of values below and above the sample statistic in a confidence interval.
We assume that the values follow normal distribution. Normal distribution is a probability that is symmetric about the mean showing the data near the mean are more frequent in occurence than data far from mean.
We know that margin of error for a confidence interval is given by:
Me=
α=1-0.95=0.05
α/2=0.025
z with α/2=1.96 (using normal distribution table)
Solving for n using formula of margin of error.
n=
=96.4
By rounding off we will get 97.
Hence the sample size required will be 97.
Learn more about standard deviation at brainly.com/question/475676
#SPJ4
The given question is incomplete and the full question is as under:
If the inspection division of a county weights and measures department wants to estimate the mean amount of soft drink fill in 2 liters bottles to within (0.01 liter with 95% confidence and also assumes that standard deviation is 0.05 liter. What is the sample size needed?
Step-by-step explanation:
A is correct option there are no real roots
because discriminant is less than zero
If X is between P and Q, then PQ is the whole line segment and XQ and XP are line segments that add up to PQ. Therefore, you can make an equation where XQ and XP add up to PQ.
XQ + XP = PQ
(3k) + (7k - 2) = 6k + 18
Combine like terms.
10k - 2 = 6k + 18
Add 2 to both sides.
10k = 6k + 20
Subtract 6k from both sides.
4k = 20
Divide both sides by 4.
k = 5
Plug 5 for k into XP.
XP = 7(5) - 2 ⇒ 35 - 2
XP = 33
