Answer: 1-91,1-42,101,161
Answer:
1.2 or 1.22 (Doesn't matter)
Step-by-step explanation:
Answer: Prokhorova is more outstanding.
Step-by-step explanation: To compare scores from different distributions, first standardize it:
z-score = 
where
x is the individual mean you want to compare
μ is the mean of the population
σ is standard deviation
For <u>Gertrud Bacher</u>:
z-score = 
z-score = 
(s)
The negative sign indicates Bacher's mean is less than the mean
For <u>Yelena Prokhorova</u>:
z-score = 
z-score = 2 (cm)
The positive sign indicates Prokhorova's mean is more than the mean.
Using z-score table, you determine the percentiles are:
For Bacher: Percentile = 5.5%
For Prokhorova: Percentile = 97.7%
Bacher's percentile means she is above 5.5% of the participants, while Prokhorova is 97.7% above the other competitors, which means Prokhorova have a better performance and deserves more points.
Answer: 4 games
Step-by-step explanation:
The team has won 6 games and lost 4 games. The win--loss ratio at the moment must be:
= 6 : 4
This must be at its simplest form which is:
= 3 : 2
To take it to 5 : 2, we would have to take up the left side of the ratio by 2.
Because we simplified by using a factor of 2, we have to multiply the 2 we just calculated by the factor to find the number of games needed to be won:
= 2 * 2
= 4 games
Answer:
p value = 0.302
Step-by-step explanation:
Given that all cars can be classified into one of four groups: the subcompact, the compact, the midsize, and the full-size. There are five cars in each group. Head injury data (in hic) for the dummies in the driver's seat are listed below.
H_0: All cars have same mean values
H_a: atleast two cars have different mean values
(Two tailed anova test)
Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
Subcompact 5 3444 688.8 48502.7
compact 5 2879 575.8 4582.7
Midsize 5 2534 506.8 18720.2
Full size 5 2689 537.8 23905.2
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 94825 3 31608.33 1.321 0.302 3.24
Within Groups 382843.2 16 23927.7
Total 477668.2 19
Since p value of 0.302 is greater than 0.05 significance level we accept null hypothesis.
p value = 0.302
