Week 8, Lab 8 cont
This deliverable continues with Z-score excercises introduced in last week’s lab
This deliverable comes from the Open Intro Statitstics Book, p. 142
Respond to the questions by answering to the post.
1. In triathlons, it is common for racers to be placed into age and gender groups.
Friends Leo and Mary both completed the Hermosa Beach Triathlon, where Leo competed in the Men, Ages
30 – 34 group while Mary competed in the Women, Ages 25 – 29 group. Leo completed the race in 1:22:28
(4948 seconds), while Mary completed the race in 1:31:53 (5513 seconds). Obviously Leo finished faster,
but they are curious about how they did within their respective groups. Can you help them? Here is some
information on the performance of their groups:
• The finishing times of the Men, Ages 30 – 34 group has a mean of 4313 seconds with a standard
deviation of 583 seconds.
• The finishing times of the Women, Ages 25 – 29 group has a mean of 5261 seconds with a standard
deviation of 807 seconds.
• The distributions of finishing times for both groups are approximately Normal.
Remember: a better performance corresponds to a faster finish.
a) Write down the short-hand for these two normal distributions.
b) What are the Z-scores for Leo’s and Mary’s finishing times? What do these Z-scores tell you?
c) Did Leo or Mary rank better in their respective groups? Explain your reasoning.
d) What percent of the triathletes did Leo finish faster than in his group?
e) What percent of the triathletes did Mary finish faster than in her group?
f) If the distributions of finishing times are not nearly normal, would your answers to parts (b) – (e) change? Explain your reasoning.
2. The average daily high temperature in June in LA is 77 ◦ F with a standard deviation of 5 ◦ F. Suppose that the temperatures in June closely follow a normal distribution.
(a) What is the probability of observing an 83 ◦ F temperature or higher in LA during a randomly chosen day in June?
(b) How cool are the coldest 10% of the days (days with lowest average high temperature) during June in LA?
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1a) Men N(μ=4313, σ=583) Women (μ=5261, σ=807)
1b) Leo Z-score = 1.08, Mary Z-score = 0.31 , these Z-scores tell us the relationship between their time and the average time.
1c) Both Mary and Leo performed well in their groups since their z-score was positive.
1d) Leo did better than 13.81% of his group.
1e) Mary did better than 37.74% of her group.
1f) Yes they would change if the distribution was not nearly normal because that could mean that there are potential outliers that could skew the data.
2a) 11.51%
2b) 70.6 degrees F
a) Men 30-34: N(μ=4313, σ=583)
Women 25-29: N(μ=5261, σ=807)
b) Mary’s z-score is 0.31
Leo’s z-score is 1.08
c) Mary did better because her z-score is smaller. This means that Mary did better in her group than Leo did in his.
d) Leo did better than 85.99% of his group.
e) Mary did better than 62.17% of her group
f) If the distribution isn’t normal then the answers will change.
2.
a) The probability of seeing a temperature higher than 83 degrees is 11.51%
b) The coldest 10% of days are 70.6 degrees.
1. a) Men: N(μ=4313, σ=583). Women: N(μ=5261, σ=807)
b) Leo: Z=1.08 and Mary: Z=0.31. These Z scores tell us the position of Leo and Mary in relation to the average finish time of each group.
c) Mary ranked better because her Z score value is lower than Leo’s.
d) Leo – 85.99%
e) Mary – 62.17%
f) No, it will not change because Z score values are in relation to other racers so the only thing changing would be the units.
2. a) 11.51%
b) 70.6
1a. Leo (μ=4313, σ=583) 4948-4313/583
Mary (μ=5261, σ=807) 5513-5261/807
1b. Leo= 1.09, Mary= .31
These z-scores tell me that Leo and Mary’s finishing times are relatively close to the mean of their respective groups.
1c. Yes, they both did well because Mary ‘s percentile score was 62.17% and Leo’s was 86.21% so they both ranked high, however Leo was the best of the two.
1d. Leo finished faster than (100-86.21) 13.79%.
1e. Mary finished faster than (100-62.17) 37.83%.
1f. I think if the distribution of finishing times was not nearly normal, the answers would change because there may be outliers or extreme values that’ll cause the data to become skewed.
2a. 11.51%
2b. 70.6 degrees fahrenheit or colder
1.
a) The short-hand normal distribution for Leo Z score is 4948-4313/583. The short-hand normal distribution for Mary Z score is 5513-5261/807.
b) The Z score for Leo’s and Mary’s finishing times are 1.09 and 0.31. These Z scores tells me that both Leo’s and Mary’s Z scores are lower and the standard deviation is also slow for the mean.
c) Yes, both Leo and Mary did better in their both respective groups because they had a number of 1.09 and a number of 0.31 that were both positive numbers and for Mary her Z score were lower.
d) Leo finished faster than 0.8621 = 86.21% . Leo had finished faster than 0.1379 = 13.79%.
e) Mary finished faster than 0.6217 = 62.17%. Mary had finished faster than 0.3783 = 37.83%.
f) If the distribution of finishing times are not nearly normal, than the answers for (b) will remain the same and for (a) and the other answers will change because the percentiles are unlike the other distributions.
2.
a) 0.1151 = 11.51%.
b) X= 70.6 during June in LA.
1.
a) The short-hand normal distribution for Leo Z score is 4948-4313/583. The short-hand normal distribution for Mary z score is 55313-5261/807.
b) The Z score for Leo’s and Mary’s finishing times are 1.09 and 0.31. These Z scores tells me that both Leo’s and Mary’s Z scores are lower and the standard deviation is also slow for the mean.
c) Yes, both Leo and Mary did better in their both respective groups because they had a number of 1.09 and a number of 0.31 that were both positive numbers and for Mary her Z score were lower.
d) Leo finished faster than 0.8621 = 86.21% . Leo had finished faster than 0.1379 = 13.79%.
e) Mary finished faster than 0.6217 = 62.17%. Mary had finished faster than 0.3783 = 37.83%.
f) If the distribution of finishing times are not nearly normal, than the answers for (b) will remain the same and for (a) and the other answers will change because the percentiles are unlike the other distributions.
2.
a) 0.1151 = 11.51%.
b) X= 70.6 during June in LA.
a). The finishing time of the age group of men between 30-34 has a normal distribution: 4313583 and 5261807. b). The z scores for Mary and Leo will be 0.31 and 1.09. c). Since Mary and Leo`s z scores are 0.31 and 1.09, it means that both are performed better in their respective groups. d). The percent of the triathletes did Leo finish faster than his group is 13.81%. e). The percent of the triathletes Mary finished faster than her group is 37.74%.
f). If the distributions of the finishing times isn`t normal, then answers will change. 2a). 0.1151.
b). X= 70.6 degrees Fahrenheit.
Z score for Leo = 4948-4313/583 = 1.089
Z score for Mary = 5513-5261/807 = 0.31
Percentile for Leo = 0.8599*100=85.99%
Percentile for Mary =0.6217*100 = 62.17%
Mary finished faster than 37.83%
Leo finished faster than 14.01%
2) 1.2
a) z= 4948-4313/583 Z=5513-5261/0807
b) Z=1.09 Z=0.31
c) leo and Mary both scored higher than the average seconds in their respective
d) 86.21%
e) 62.17%
f)
2.
a) 1.2 this shows its higher
b) 34.51%