Respuesta :
Answer:
a) 0.5 = 50% probability that an elite athlete has a maximum oxygen uptake of at least 70 ml/kg
b) 0.3050 = 30.50% probability that an elite athlete has a maximum oxygen uptake of 65 ml/kg or lower
c) NO.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Z scores greater than 2 or lower than -2 means that X is considered unlikely.
In this problem, we have that:
[tex]\mu = 70, \sigma = 9.8[/tex]
(a) What is the probability that an elite athlete has a maximum oxygen uptake of at least 70 ml/kg?
This is 1 subtracted by the pvalue of Z when X = 70. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{70 - 70}{9.8}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] has a pvalue of 0.5.
1 - 0.5 = 0.5
0.5 = 50% probability that an elite athlete has a maximum oxygen uptake of at least 70 ml/kg
b) What is the probability that an elite athlete has a maximum oxygen uptake of 65 ml/kg or lower?
This is the pvalue of Z when X = 65. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{65 - 70}{9.8}[/tex]
[tex]Z = -0.51[/tex]
[tex]Z = -0.51[/tex] has a pvalue of 0.3050.
0.3050 = 30.50% probability that an elite athlete has a maximum oxygen uptake of 65 ml/kg or lower
(c) Consider someone with a maximum oxygen uptake of 38 ml/kg. Is it likely that this person is an elite athlete?
We have to find the z-score. If it is lower than -2, we write no.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{38 - 70}{9.8}[/tex]
[tex]Z = -3.27[/tex]
So the answer is NO.
The probability will be:
(a) 0.5
(b) 0.3050
(c) No, person isn't an elite athlete.
According to the question,
Mean,
- [tex]\mu = 70[/tex]
Standard deviation,
- [tex]\sigma = 9.8[/tex]
(a)
Maximum oxygen uptake,
- X = 70
The probability is:
→ [tex]Z = \frac{X- \mu}{\sigma}[/tex]
[tex]= \frac{70-70}{9.8}[/tex]
[tex]= 0[/tex] (p-value of 0.5)
then,
[tex]= 0.5[/tex] or 50%
(b)
- X = 65
→ [tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]= \frac{65-70}{9.8}[/tex]
[tex]= -0.51[/tex] (p-value of 0.3050)
then,
[tex]= 0.3050[/tex] or, 30.50%
(c)
If it is lower than "-2", then we have to find the z-score.
→ [tex]Z = \frac{X-\mu}{\sigma}[/tex]
[tex]= \frac{38-70}{9.8}[/tex]
[tex]= -3.27[/tex]
Thus the above approach is appropriate.
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