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. What percentage 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?

Answers

Answer 1

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

Answer 2

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


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у
Gas
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0
0
Distance (miles)
350
300
250
200
150
100
50
60
vlund
100
140
X
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1 2 3 4 5 6 7 8 9 10 11
Gas (gallons)
10
200
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