A high school basketball coach is selecting his team. The minimum and maximum height requirements are as follows:

Point Guard: Small Forward: Center:

6’2" – 6’6" tall at least 6’6" tall at least 6’8" tall

The heights of the Lincoln High School boys have a normal distribution with a mean height of 70 inches and the standard deviation of 4 inches.

Which statement is true?

A. About 16% of the boys are eligible to be a point guard on the team. B. About 2% of the boys are eligible to be a small forward on the team. C. About 48% of the boys are eligible to be a small forward on the team. D. About 34% of the boys are eligible to be a point guard on the team.

Answer:

B. About 2% of the boys are eligible to be a small forward on the team

Step-by-step explanation:

Recall : 1 feets = 12 inches

Point guard = 6’2" – 6’6" tall = 74 - 78 inches

Mean = 70 ; Standard deviation = 4

Z = (x - mean) / standard

P(x < 74) = (74 - 70) / 4 = 1

P(x < 78) = (78 - 70) / 4 = 2

0.97725 - 0.84134 = 0.13591

Small forward : 6'6" = 78 inches

P(x ≥ 78) = (78 - 70) / 4 = 2

P(z ≥ 2) = 0.02275 = 2.275% about 2%

Centre : 6'8" = 80

P(x ≥ 80) = (80 - 70) / 4 = 2.5

P(z ≥ 2.5) = 0.0062097 = 0.62%

About 2% of the boys are eligible to be a small forward on the team. Then the correct option is B.

The z-score is a numerical measurement used in statistics of the value's relationship to the mean of a group of values, measured in terms of standards from the mean.

A high school basketball coach is selecting his team.

The minimum and maximum height requirements are as follows:

Point Guard: Small Forward: Center:

6’2" – 6’6" tall at least 6’6" tall at least 6’8" tall

The heights of the Lincoln High School boys have a normal distribution with a mean height of 70 inches and a standard deviation of 4 inches.

The value of the z-score will be

[tex]z = \dfrac{x - \mu }{\sigma}[/tex]

[tex]P(x < 74) = \dfrac{74-70}{4} = 1 \\\\P(x < 78) = \dfrac{78-70}{4} = 2[/tex]

0.97725 - 0.84134 = 0.13591

Small forward: 6'6" = 78 inches

[tex]P(x \geq 78) = \dfrac{78- 70 }{4} 2\\\\\\P(z \geq 2) = 0.02275 = 2.275 \%[/tex]

Centre: 6'8" = 80 inches

[tex]P(x \geq 80) = \dfrac{80- 70 }{4} 2\\\\\\P(z \geq 2.5) = 0.0062097= 0.62 \%[/tex]

More about the z-score link is given below.

https://brainly.com/question/13299273

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