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2 years ago

14

240 people are going to a charity event. 3/5 of the guests have ordered chicken for their meals. Of the remaining guests, 12.5%

have ordered gluten free meals.
How many people haven't ordered their meals yet??

1 answer:

fiasKO [112]2 years ago

5 0

Answer:

84

Step-by-step explanation:

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A random variable x follows a normal distribution with mean d and standard deviation o=2. It is known that x is less than 5 abou

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The mean of this distribution is approximately 3.96.

Step-by-step explanation:

Here's how to solve this problem using a normal distribution table.

Let $z$ be the

$\displaystyle z = \frac{x - \mu}{\sigma}$ .

In this question, $x = 5$ and $\sigma = 2$ . The equation becomes

$\displaystyle z = \frac{5 - \mu}{2}$ .

To solve for $\mu$ , the mean of this distribution, the only thing that needs to be found is the value of $z$ . Since

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The problem is that the normal distribution tables list only the value of $P(0 \le Z \le z)$ for $z \ge 0$ . To estimate $z$ from $P(Z \le z) = 0.6985$ , it would be necessary to find the appropriate

Since $P(Z \le z) = 0.6985$ and is greater than $P(Z \le 0) = 0.50$ , $z > 0$ . As a result, $P(Z \le z)$ can be written as the sum of $P(Z < 0)$ and $P(0 \le Z \le z)$ . Besides, $P(Z < 0) = P(Z \le 0) = 0.50$ . As a result:

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Therefore:

$\begin{aligned}&P(0 \le Z \le z) \\ &= P(Z \le z) - 0.50 \\&= 0.6985 - 0.50 \\&=0.1985 \end{aligned}$ .

Lookup $0.1985$ on a normal distribution table. The corresponding $z$ -score is $0.52$ . (In other words, $P(0 \le Z \le 0.52) = 0.1985$ .)

Given that

$z = 0.52$ ,
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$\displaystyle 0.52 = \frac{5 - \mu}{2}$ .

$\mu = 5 - 2 \times 0.52 = 3.96$ .

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