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2 years ago

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Assume that the heights of boys in a high school basketball tournament are normally distributed, with mean 70 inches and standar

d deviation 2.5 inches. What is the expected number of boys in a group of 40 who are taller than 70 inches?

1 answer:

Annette [7]2 years ago

6 0

Answer: 20

Step-by-step explanation:

We assume that the heights of boys in a high school basketball tournament are normally distributed.

Given : Mean height of boys : $\mu=70$ inches.

Standard deviation: $\sigma=2.5$ inches.

Let x denotes the heights of boys in a high school basketball tournament .

Then the probability that a boy is taller than 70 inches will be :-

$P(x> 70)=1-P(x\leq70)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{70-70}{2.5})\\\\=1-P(z\leq0)=1-0.5=0.5\ \ \text{[by using z-value table]}$

Now, the expected number of boys in a group of 40 who are taller than 70 inches will be :-

$40\times0.5=20$

Hence, the expected number of boys in a group of 40 who are taller than 70 inches=20

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Binomial probability distribution

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$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$n = 14, p = 0.5$

Compute the probability of selling more than 4 properties in one week.

Either you sell 4 or less properties in one week, or you sell more. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 4) + P(X > 4) = 1$

We want to find $P(X > 4)$ . So

$P(X > 4) = 1 - P(X \leq 4)$

In which

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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$P(X = 3) = C_{14,3}.(0.5)^{3}.(0.5)^{11} = 0.0222$

$P(X = 4) = C_{14,4}.(0.5)^{4}.(0.5)^{10} = 0.0611$

So

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91.02% probability of selling more than 4 properties in one week.

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