Answer:
a) We need to find p(x<68.15)
Standardize x to z, we have, z = (68.15-μ)/σ
=(68.15-57)/8.1
=1.38
Reading from standard normal tables,
p(z<1.38) =0.9162.
b) We need to find p(x>54.4)
Standardize x to z, we have, z = (54.4-μ)/σ
=(54.4-57)/8.1
=-0.32
Reading from standard normal tables,
p(z>-0.32) = 0.6255
Nidyaj backwards ny is the real name
Answer:
I used the function 1-proportion z-interval on the calculator, where I inputted
- Successes(x) = 64
- Sample size(n) = 593
- Confidence Level(C level) = 0.99
It would result in zInterval_1Prop 64,593,0.99: stat.results, where the values are shown below as:
- Lower bound (CLower) = 0.075105
- Upper bound (CUpper) = 0.140747
- test statistic (p^) = 0.107926
- Margin of error (ME) = 0.032821
Therefore, the 99% confidence interval would be around 0.11 ± 0.03 or range from 0.08 to 0.14.
<em><u>Note: not sure if this is correct</u></em><em> O_o</em>
A. providing letters of recommendation for colleges. ~~~APEX
Answer:
search up previous benchmarks online, thats a good way to study
Explanation:
