Respuesta :
Answer:
a) 89.97% of men have arm spans between 66 and 76 inches.
b) The z-score for this person's arm span is 5.68. 0% of males have an arm span at least as long as this person
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be 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.
Mean of 71.4 inches and a standard deviation of 3.1 inches.
This means that [tex]\mu = 71.4, \sigma = 3.1[/tex]
a. What percentage of men have arm spans between 66 and 76 inches?
The proportion is the pvalue of Z when X = 76 subtracted by the pvalue of Z when X = 66. The percentage is the proportion multiplied by 100.
X = 76
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{76 - 71.4}{3.1}[/tex]
[tex]Z = 1.48[/tex]
[tex]Z = 1.48[/tex] has a pvalue of 0.9306
X = 66
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{66 - 71.4}{3.1}[/tex]
[tex]Z = -1.74[/tex]
[tex]Z = -1.74[/tex] has a pvalue of 0.0409
0.9306 - 0.0409 = 0.8997
0.8997*100% = 89.97%
89.97% of men have arm spans between 66 and 76 inches.
b. A particular professional basketball player has an arm span of almost 89 inches. Find the z-score for this person's arm span. What percentage of males have an arm span at least as long as this person?
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{89 - 71.4}{3.1}[/tex]
[tex]Z = 5.68[/tex]
The z-score for this person's arm span is 5.68.
Z = 5.68 has a pvalue of 1
1 - 1 = 0
0% of males have an arm span at least as long as this person
The percentage of men who have arm spans between 66 and 76 inches is 89.97% and the percentage of males who have an arm span at least as long as this person is 0%.
Given :
According to survey data, the distribution of arm spans for males is approximately Normal with a mean of 71.4 inches and a standard deviation of 3.1 inches.
a) The formula of z-score is given below:
[tex]\rm Z = \dfrac{X-\mu}{\sigma}[/tex]
where [tex]\mu[/tex] is the mean and [tex]\sigma[/tex] is the standard deviation.
When X = 66 the formula of z-score becomes:
[tex]\rm Z = \dfrac{66-71.4}{3.1}[/tex]
Z = -1.74
The p-value regarding the (Z = -1.74) is 0.0409.
When X = 76 the formula of z-score becomes:
[tex]\rm Z = \dfrac{76-71.4}{3.1}[/tex]
Z = 1.48
The p-value regarding the (Z = 1.48) is 0.9306.
Now, the percentage of men who have arm spans between 66 and 76 inches is:
= (0.9306 - 0.0409) [tex]\times[/tex] 100 %
= 89.97%
b) Now, at X = 89 the formula of z-score becomes:
[tex]\rm Z = \dfrac{89-71.4}{3.1}[/tex]
Z = 5.68
The p-value regarding the (Z = 5.68) is 1.
So, the percentage of males who have an arm span at least as long as 89 inches is:
= (1 - 1) [tex]\times[/tex] 100 %
= 0%
For more information, refer to the link given below:
https://brainly.com/question/13299273
