Math / Data & Statistics

QuestionLet $X$ be normally distributed with mean $\mu=10$ and standard deviation $\sigma=6$ . Note: You may find it useful to reference the $z$ table. a. Find $P(X \leq 0)$ .
Note: Round your final answer to 4 decimal places. $P(X \leq 0)$ b. Find $P(X>2)$ .
Note: Round your final answer to 4 decimal places. $P(X>2)$

Studdy Solution

STEP 1

1. $X$ is a normally distributed random variable with mean $\mu = 10$ and standard deviation $\sigma = 6$ .
2. The standard normal distribution table (z-table) will be used to find probabilities.
3. Probabilities will be rounded to four decimal places.

STEP 2

1. Calculate $P(X \leq 0)$ .
2. Calculate $P(X > 2)$ .

STEP 3

Convert the random variable $X$ to a standard normal variable $Z$ using the formula:
$Z = \frac{X - \mu}{\sigma}$
For $X = 0$ :
$Z = \frac{0 - 10}{6} = \frac{-10}{6} = -\frac{5}{3} \approx -1.67$

STEP 4

Use the z-table to find $P(Z \leq -1.67)$ .
From the z-table:
$P(Z \leq -1.67) \approx 0.0475$
Thus, $P(X \leq 0) \approx 0.0475$ .

STEP 5

Convert the random variable $X$ to a standard normal variable $Z$ for $X = 2$ :
$Z = \frac{2 - 10}{6} = \frac{-8}{6} = -\frac{4}{3} \approx -1.33$

STEP 6

Use the z-table to find $P(Z \leq -1.33)$ .
From the z-table:
$P(Z \leq -1.33) \approx 0.0918$
Thus, $P(X \leq 2) \approx 0.0918$ .

STEP 7

Calculate $P(X > 2)$ using the complement rule:
$P(X > 2) = 1 - P(X \leq 2)$
$P(X > 2) = 1 - 0.0918 = 0.9082$
The probabilities are:
a. $P(X \leq 0) \approx \boxed{0.0475}$
b. $P(X > 2) \approx \boxed{0.9082}$

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Studdy Solution

STEP 1

1. $X$ is a normally distributed random variable with mean $\mu = 10$ and standard deviation $\sigma = 6$ .
2. The standard normal distribution table (z-table) will be used to find probabilities.
3. Probabilities will be rounded to four decimal places.

STEP 2

1. Calculate $P(X \leq 0)$ .
2. Calculate $P(X > 2)$ .

STEP 3

Convert the random variable $X$ to a standard normal variable $Z$ using the formula:
$Z = \frac{X - \mu}{\sigma}$
For $X = 0$ :
$Z = \frac{0 - 10}{6} = \frac{-10}{6} = -\frac{5}{3} \approx -1.67$

STEP 4

Use the z-table to find $P(Z \leq -1.67)$ .
From the z-table:
$P(Z \leq -1.67) \approx 0.0475$
Thus, $P(X \leq 0) \approx 0.0475$ .

STEP 5

Convert the random variable $X$ to a standard normal variable $Z$ for $X = 2$ :
$Z = \frac{2 - 10}{6} = \frac{-8}{6} = -\frac{4}{3} \approx -1.33$

STEP 6

Use the z-table to find $P(Z \leq -1.33)$ .
From the z-table:
$P(Z \leq -1.33) \approx 0.0918$
Thus, $P(X \leq 2) \approx 0.0918$ .

STEP 7

Calculate $P(X > 2)$ using the complement rule:
$P(X > 2) = 1 - P(X \leq 2)$
$P(X > 2) = 1 - 0.0918 = 0.9082$
The probabilities are:
a. $P(X \leq 0) \approx \boxed{0.0475}$
b. $P(X > 2) \approx \boxed{0.9082}$

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Studdy Solution

STEP 1

1. X X X is a normally distributed random variable with mean μ=10 \mu = 10 μ=10 and standard deviation σ=6 \sigma = 6 σ=6.2. The standard normal distribution table (z-table) will be used to find probabilities.3. Probabilities will be rounded to four decimal places.

STEP 2

1. Calculate P(X≤0) P(X \leq 0) P(X≤0).2. Calculate P(X>2) P(X > 2) P(X>2).

STEP 3

Convert the random variable X X X to a standard normal variable Z Z Z using the formula:Z=X−μσ Z = \frac{X - \mu}{\sigma} Z=σX−μ​For X=0 X = 0 X=0:Z=0−106=−106=−53≈−1.67 Z = \frac{0 - 10}{6} = \frac{-10}{6} = -\frac{5}{3} \approx -1.67 Z=60−10​=6−10​=−35​≈−1.67

STEP 4

Use the z-table to find P(Z≤−1.67) P(Z \leq -1.67) P(Z≤−1.67).From the z-table:P(Z≤−1.67)≈0.0475 P(Z \leq -1.67) \approx 0.0475 P(Z≤−1.67)≈0.0475Thus, P(X≤0)≈0.0475 P(X \leq 0) \approx 0.0475 P(X≤0)≈0.0475.

STEP 5

Convert the random variable X X X to a standard normal variable Z Z Z for X=2 X = 2 X=2:Z=2−106=−86=−43≈−1.33 Z = \frac{2 - 10}{6} = \frac{-8}{6} = -\frac{4}{3} \approx -1.33 Z=62−10​=6−8​=−34​≈−1.33

STEP 6

Use the z-table to find P(Z≤−1.33) P(Z \leq -1.33) P(Z≤−1.33).From the z-table:P(Z≤−1.33)≈0.0918 P(Z \leq -1.33) \approx 0.0918 P(Z≤−1.33)≈0.0918Thus, P(X≤2)≈0.0918 P(X \leq 2) \approx 0.0918 P(X≤2)≈0.0918.

STEP 7

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1. $X$ is a normally distributed random variable with mean $\mu = 10$ and standard deviation $\sigma = 6$ .
2. The standard normal distribution table (z-table) will be used to find probabilities.
3. Probabilities will be rounded to four decimal places.

1. Calculate $P(X \leq 0)$ .
2. Calculate $P(X > 2)$ .

Convert the random variable $X$ to a standard normal variable $Z$ using the formula:
$Z = \frac{X - \mu}{\sigma}$
For $X = 0$ :
$Z = \frac{0 - 10}{6} = \frac{-10}{6} = -\frac{5}{3} \approx -1.67$

Use the z-table to find $P(Z \leq -1.67)$ .
From the z-table:
$P(Z \leq -1.67) \approx 0.0475$
Thus, $P(X \leq 0) \approx 0.0475$ .

Convert the random variable $X$ to a standard normal variable $Z$ for $X = 2$ :
$Z = \frac{2 - 10}{6} = \frac{-8}{6} = -\frac{4}{3} \approx -1.33$

Use the z-table to find $P(Z \leq -1.33)$ .
From the z-table:
$P(Z \leq -1.33) \approx 0.0918$
Thus, $P(X \leq 2) \approx 0.0918$ .