Answer:
Therefore the answer is 20.
Step-by-step explanation:
We know that
class interval = range / number of classes
But here number of classes is not given , so we use the formula
class interval = range / ( 1+ 3.322 log N)
where , range =maximum - minimum = 220-100 = 120
N= number of observations = 50
class interval = 120 / ( 1+ 3.322 * log 50) = 18.06
Rounding up to a convinient number
Thus , class intervai = 20
Therefore the answer is 20.
Answer:
yes I have an idea jnhvhbvcf
Answer: A
Step-by-step explanation:
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<h3>Andre Bought:</h3><h3>• 1 Baseball Glove For $34 Each</h3><h3>• 2 Packs of Socks For $6 - 7.75% Each</h3>
<h3 /><h3>$34 + $8</h3><h3>= $42</h3><h3><u>Andre Paid A Total of $42</u></h3>
<h3>7.75% × $6 ÷ 100</h3><h3>= 0.465 × 2</h3><h3>= 0.93%</h3><h3><u>Andre Saved 0.93% In Total At The Sale</u></h3>
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Answer:
(0.5848 ; 0.6552)
We are confident that about 58% to 66% of sea foods in the country are Mislabelled.
No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.
Step-by-step explanation:
Sample size, n = 51
p = 0.62
1 - p = 1 - 0.62 = 0.38
n = 515
Confidence level = 90% = Zcritical at 90% = 1.645
Confidence interval = (p ± margin of error)
Margin of Error = Zcritical * sqrt[(p(1-p))/n]
Margin of Error = 1.645 * sqrt[(0.62(0.38))/515]
Margin of Error = 1.645 * 0.0214
Margin of Error = 0.035203
Lower boundary = (0.62 - 0.035203) = 0.584797
Upper boundary = (0.62 + 0.035203) = 0.655203
(0.5848 ; 0.6552)
We are confident that about 58% to 66% of sea foods in the country are Mislabelled.
No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.
