Answer:
The probability that the average age of a randomly selected sample of 100 students will be less than 21.8 years is 0.159
Step-by-step explanation:
According to the given data we have the following:
mean = μ= 22
standard deviation = σ = 2
n = 100
μx = 22
σx=σ/√n=2/√100=0.2
Therefore, P( x < 21.8)=P(x-μx)/σx<(21.8-22)/0.2
=P(z<-1)
= 0.159
The probability that the average age of a randomly selected sample of 100 students will be less than 21.8 years is 0.159
Answer:
not statistically significant at ∝ = 0.05
Step-by-step explanation:
Sample size( n ) = 61
Average for student leader graduates to finish degree ( x') = 4.97 years
std = 1.23
Average for student body = 4.56 years
<u>Determine if the difference between the student leaders and the entire student population is statistically significant at alpha</u>
H0( null hypothesis ) : u = 4.56
Ha : u ≠ 4.56
using test statistic
test statistic ; t = ( x' - u ) / std√ n
= ( 4.97 - 4.56 ) / 1.23 √ 61
= 2.60
let ∝ = 0.05 , critical value = -2.60 + 2.60
Hence we wont fail to accept H0
This shows that the difference between the student leaders and the entire student population is not statistically significant at ∝ = 0.05
<span>(38-63)×6÷Z
=</span><span>(38-63)×6÷84
=-25</span>×6÷84
=150
=1 66/84
= 1 33/42
The smaller angle = 58
the larger angle = 122
reply to me in comments if you want the proof
Answer:
8.2
Step-by-step explanation:
4(4)-7.8
