In a standard normal distribution, what z value corresponds to 17% of the data between the mean and the z value?

1 answer:

hjlf1 year ago

4 0

You're looking for a value $z$ such that

$\mathbb P(0$

Because the distribution is symmetric, the value of $z$ in either case will be the same.

Now, because the distribution is continuous, you have that

$0.17=\mathbb P(0$

The mean for the standard normal distribution is $0$ , and because the distribution is symmetric about its mean, it follows that $\mathbb P(Z$ .

$0.17=\mathbb P(Z$

You can consult a $z$ score table to find the corresponding score for this probability. It turns out to be $z\approx0.4399$ .

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Peanut allergies among children. About 2% of children in the United States are allergic to peanuts. Choose three children at ran

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Answer:

P(X=2|X≥1) = 0.0199973.

Step-by-step explanation:

The chance that each of the three children is allergic to peanuts is independent. Each with a chance of 2% = 0.02.

X follows a binomial distribution with

3 trials, and
a chance of "success" of 2% = 0.02.

$P(X = 2|X\ge 1) = \dfrac{P(X = 2 \cap X\ge 1)}{P(X\ge 1)}$ .

The event $X \ge 1$ includes

$X = 1$
${\bf X = 2}$ , and
$X = 3$ .

In other words, $X = 2$ implies that $X\ge 1$ .

$P(X = 2 \cap X\ge 1) = P(X = 2) = \left(\begin{array}{c}3\\2\end{array}\right ) \times 0.02^{2} \times (1-0.02) = 0.001176$ .

$P(X\ge 1) = 1 - P(X< 1) \\\phantom{P(X\ge 1)}= 1 - P(X = 0) \\\phantom{P(X\ge 1)}= 1 -\left(\begin{array}{c}3\\0\end{array}\right ) \times (1-0.02)^{3} \\\phantom{P(X\ge 1)}= 0.058808$ .