An instructor asked her students how much time (to the nearest hour) they spent studying for the midterm. The data are displayed
1 answer:
Based on the data displayed in the histogram, the percentage of students who study 2 or more hours for the midterm is 88%.
<h3>What percentage study 2 hours or more?</h3>First, find the number of students studying:
= 3 + 9 + 6 + 3 + 2 + 1 + 1
= 25
The number of students studying 2 or more hours for the exam are:
= 9 + 6 + 3 + 2 + 1 + 1
= 22 students
The percentage studying for two hours or more is:
= 22 / 25
= 88%
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Answer:
95% confidence interval for the population is; -0.9525 < < -0.9505
Step-by-step explanation:
Given the data in the question;
correlation coefficient y between output and time under a load of 588 N was −0.9515
so first we determine the quantity W.
we know that; W = ln(
)
we substitute -0.9515 for r
W = ln(
)
W = ln(
)
W = ln(
)
W = ln( 0.02485 )
W = ( -3.6948975 )
W = - 1.8474
So the quantity W is normally distributed with standard deviation given by;
σ = 1 / √(n - 3)
given that n is 33,000, we substitute
σ = 1 / √(33,000 - 3)
σ = 1 / √32997
σ = 0.0055
Now, at 95% confidence interval, μ will be;
⇒ - 1.8474 - 1.96( 0.0055 ) < μ < - 1.8474 + 1.96( 0.0055 )
⇒ - 1.8474 - 0.01078 < μ < - 1.8474 + 0.01078
⇒ -1.8582 < μ < -1.8366
So to obtain 95% confidence interval for p, we use the following equation;
we transform the inequality
p = [( ) / (
)]
we substitute
<
<
<
<
<
<
<
<
-0.9525 < < -0.9505
Therefore, 95% confidence interval for the population is; -0.9525 < < -0.9505