Find the surface area of the square pyramid using its net

The weekly amount spent by a small company for in-state travel has approximately a normal distribution with mean $1450 and stand

Llana [10]

Answer:

0.0903

Step-by-step explanation:

Given that :

The mean = 1450

The standard deviation = 220

sample mean = 1560

$P(X > 1560) = P( Z > \dfrac{x - \mu}{\sigma})$

$P(X > 1560) = P(Z > \dfrac{1560 - 1450}{220})$

$P(X > 1560) = P(Z > \dfrac{110}{220})$

P(X> 1560) = P(Z > 0.5)

P(X> 1560) = 1 - P(Z < 0.5)

From the z tables;

P(X> 1560) = 1 - 0.6915

P(X> 1560) = 0.3085

Let consider the given number of weeks = 52

Mean $\mu_x$ = np = 52 × 0.3085 = 16.042

The standard deviation = $\sqrt {n \time p (1-p)}$

The standard deviation = $\sqrt {52 \times 0.3085 (1-0.3085)}$

The standard deviation = 3.3306

Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.

Then;

Pr ( Y > 20) = P( z > 20)

$Pr ( Y > 20) = P(Z > \dfrac{20.5 - 16.042}{3.3306})$

$Pr ( Y > 20) = P(Z >1 .338)$

From z tables

P(Y > 20) $\simeq$ 0.0903

7 0

3 years ago

018 10.0 points

Elina [12.6K]

Answer:

what grade

Step-by-step explanation:

7 0

3 years ago

3. The adult men of the Dinaric Alps have the highest average height of all regions. The

ahrayia [7]

Using the normal distribution, it is found that:

3 - a) The 40th percentile of the height of Dinaric Alps distribution for men is of 72.2 inches.

3 - b) The minimum height of man in the Dinaric Alps that would place him in the top 10% of all heights is of 76.84 inches.

4 - a) The 25th percentile for the math scores was of 71.6 inches.

4 - b) The 75th percentile for the math scores was of 78.4 inches.

<h3>Normal Probability Distribution </h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

Question 3:

The mean is of 73 inches, hence $\mu = 73$ .

The standard deviation is of 3 inches, hence $\sigma = 3$ .

Item a:

The 40th percentile is X when Z has a p-value of 0.4, so X when Z = -0.253.

$Z = \frac{X - \mu}{\sigma}$

$-0.253 = \frac{X - 73}{3}$

$X - 73 = -0.253(3)$

$X = 72.2$

The 40th percentile of the height of Dinaric Alps distribution for men is of 72.2 inches.

Item b:

The minimum height is the 100 - 10 = 90th percentile is X when Z has a p-value of 0.9, so X when Z = 1.28.

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 73}{3}$

$X - 73 = 1.28(3)$

$X = 76.84$

The minimum height of man in the Dinaric Alps that would place him in the top 10% of all heights is of 76.84 inches.

Question 4:

The mean score is of 75, hence $\mu = 75$ .

The standard deviation is of 5, hence $\sigma = 5$ .

Item a:

The 25th percentile is X when Z has a p-value of 0.25, so X when Z = -0.675.

$Z = \frac{X - \mu}{\sigma}$

$-0.675 = \frac{X - 75}{5}$

$X - 75 = -0.675(5)$

$X = 71.6$

The 25th percentile for the math scores was of 71.6 inches.

Item b:

The 75th percentile is X when Z has a p-value of 0.25, so X when Z = 0.675.

$Z = \frac{X - \mu}{\sigma}$

$0.675 = \frac{X - 75}{5}$

$X - 75 = 0.675(5)$

$X = 78.4$

The 75th percentile for the math scores was of 78.4 inches.

To learn more about the normal distribution, you can take a look at brainly.com/question/24663213

5 0

2 years ago

Help me solve y-(-2)-1

Maslowich

The answer would be y+1

3 0

3 years ago

Find the product of (√3-5) and (√3+2)

beks73 [17]

Step-by-step explanation:

check it.. i have solved it for you.

3 0

2 years ago