Question

In which of the following situations would it be appropriate to use a Normal distribution to approximate probabilities for a binomial distribution with the given values of n and p?
(a) n = 10, p = 0.5
(b) n = 40, p = 0.88
(c) n = 100, p = 0.2
(d) n = 100, p = 0.99
(e) n = 1000, p = 0.003

Answer 1

Answer:

(c) n = 100, p = 0.2

Step-by-step explanation:

To approximate to use the normal distribution for a binomial, we need that:

$np \geq 10, n(1-p)\geq 10$

So, for each option

(a) n = 10, p = 0.5

$np = 10*0.5 = 5 < 10$

So not appropriate

(b) n = 40, p = 0.88

$np = 40*0.88 = 35.2 > 10$

$n(1-p) = 40*0.12 = 4.8 < 10$

So not appropriate.

(c) n = 100, p = 0.2

$np = 100*0.2 = 20 > 10$

$n(1-p) = 100*0.8 = 80 > 10$

So appropriate

(d) n = 100, p = 0.99

$np = 100*0.99 = 99 > 10$

$n(1-p) = 100*0.01 = 1 < 10$

So not appropriated

(e) n = 1000, p = 0.003

$np = 1000*0.003 = 3 < 10$

So not appropriate

Answer 2

A Normal distribution to approximate probabilities for a binomial distribution with the given values of n = 100 and p = 0.2.

Given that,

Normal distribution to approximate probabilities for a binomial distribution with the given values of n and p.

We have to determine,

In which of the following situations would it be appropriate to use a normal distribution to approximate probabilities for a binomial distribution with the given values of n and p.

According to the question,

Probabilities for a binomial distribution with the given values of n and p.

(a) n = 10, p = 0.5

(b) n = 40, p = 0.88

(c) n = 100, p = 0.2

(d) n = 100, p = 0.99

(e) n = 1000, p = 0.003

To approximate to use the normal distribution for a binomial is given as,

$np\geq 10 , \ n(1-p)\geq 10$

• When n = 10 and p = 0.5,

        $= np \geq 10\\\\= 10 \times 0.5 = 5<10$

The normal distribution is less than 10 so, this not appropriate.

• When n = 40 and p = 0.88,

        $= n.p \geq 10\\\\= (40).(0.88) \geq 10\\\\= 35.2\geq 10\\\\Then,\\=n(1-p) = 40(1-0.88) \geq 10 \\\\= 40\times 0.12 \geq 10 = 4.8\leq 10$

The normal distribution is less than 10 so, this not appropriate.

• When n = 100 and p =0.2,

        $= n.p \geq 10\\\\= (100).(0.2) \geq 10\\\\= 20\geq 10\\\\Then,\\=n(1-p) = 100(1-0.2) \geq 10 \\\\= 100\times 0.8\geq 10 = 80\geq 10$

The normal distribution is greater than 10 so, this appropriate.

• When n = 100 and p = 0.99,

        $= n.p \geq 10\\\\= (100).(0.99) \geq 10\\\\= 99\geq 10\\\\Then,\\=n(1-p) = 100(1-0.99) \geq 10 \\\\= 100\times 0.01\geq 10 = 1\leq 10$

The normal distribution is less than 10 so, this not appropriate.

• When n = 1000 and p = 0.003,

        $= np \geq 10\\\\= 1000 \times 0.003 = 3<10$

The normal distribution is less than 10 so, this not appropriate.

Hence, A Normal distribution to approximate probabilities for a binomial distribution with the given values of n = 100 and p = 0.2.

To know more about Binomial Distribution click the link given below.

https://brainly.com/question/16776778