Answer: One hundred thousand
Step-by-step explanation:
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Complete question :
The 100m dah times in the girl's track meet were normally distributed with a mean of 13 seconds and a standard deviation of 0.3 seconds.
Lana finished the race in 13.2 seconds . If 84 other girls ran in the event, approximately how many runners did she beat?
Answer:
21 runners
Step-by-step explanation:
Mean, μ = 13 seconds
Standard deviation, σ = 0.3
Lana's race time = 13.2
We find the proportion of runners who had race time above 13.2 ;
Proportion of who had race tune above 13.2
P(x > 13.2)
Obtain the Zscore
Zscore = (x - μ) / σ
Z = (13.2 - 13) / 0.3
Zscore = 0.2 / 0.3 = 0.6667
P(Z > 0.6667) = 0.25239 (Z probability calculator)
This is about 0.25239 * 100 = 25.24% = 25% (nearest percent)
Hence, Number of runners Lana beat = 25% of total runners ;
0.25 * 84 = 21
Hence, Lana beat about 21 runners
Note that two negatives = one positive
- (-4 1/3) = + 4 1/3
Subtract
-9 + 4 1/3 = -4 2/3
-4 2/3 is your answer
hope this helps
Answer:
With 95% confidence when n equals 44, the population mean is between a lower limit of 33.96 and an upper limit of 40.04
Step-by-step explanation:
Confidence Interval = mean + or - Error margin (E)
mean = 37
sd = 10
n = 44
degree of freedom = n - 1 = 44 - 1 = 43
t-value corresponding to 43 degrees of freedom and 95% confidence level is 2.0165
E = t × sd/√n = 2.0165×10/√44 = 3.04
Lower limit = mean - E = 37 - 3.04 = 33.96
Upper limit = mean + E = 37 + 3.04 = 40.04
95% confidence interval is between 33.96 and 40.04
Answer:
Exact form 9/4 Mixed number form 2 1/4 decimal form 2.25
Step-by-step explanation:
