What was the name of Jupiter’s big pet dog

Suppose that the weight of seedless watermelons is normally distributed with mean 6.1 kg. and standard deviation 1.9 kg. Let X b

horrorfan [7]

Answer:

a. $X\sim N(\mu = 6.1, \sigma = 1.9)$ b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349

Step-by-step explanation:

a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.

b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.

c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842

d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166

e. The z-score related to 6.4 kg is $z_{1} = (6.4-6.1)/1.9 = 0.1579$ and the z-score related to 7 kg is $z_{2} = (7-6.1)/1.9 = 0.4737$ , we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194

f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349

7 0

2 years ago

Tire pressure monitoring systems (TPMS) warn the driver when the tire pressure of the vehicle is 28% below the target pressure.

Alona [7]

Answer:

Step-by-step explanation:

Hello!

The monitoring system warn the driver when the tire pressure of the vehicle is 28% below target pressure.

Be X: target tire pressure of a certain car (pounds per square inch)

a)

X= 28 psi

If the monitoring system will warn the driver when the pressure is 28% below the target pressure: X-0.28X

First step, you have to calculate the 28% of 28psi

28*0.28= 7.84

Second step, is to subtract the calculated 28% to the target pressure:

28 - 7.84= 20.16

The TPMS will trigger a warning at 20.16 psi.

b)

If X~N(μ;σ²)

μ= 28psi (since the average is on target, then the target pressure for the car will be the average value of the distribution)

σ= 3psi

P(X≤20.16)

The standard normal distribution is tabulated. Any value of any random variable X with normal distribution can be "converted" by subtracting the variable from its mean and dividing it by its standard deviation.

So to calculate each of the asked probabilities, you have to first, "transform" the value of the variable to a value of the standard normal distribution Z, then you use the standard normal tables to reach the corresponding probability.

Z= (X-μ)/σ= (20.16-28)/3= -2.61

Now you have to look for the corresponding value of probability using the Z-table. Since the value is negative you have to the use the left entry of the Z-table, in the first column you'll find the integer and first decimal of the value -2.6- and in the first row you'll find the second decimal value -.-1

The value of probability that corresponds to -2.61 is:

P(Z≤-2.61)= 0.005

c)

You have to calculate the probability of inspecting a tire at random and it being inflated within recommended range, symbolically this is:

P(30≤X≤26)= P(X≤30)-P(X≤26)

Calculate both Z values:

Z= (30-28)/3= 0.67

Z= (26-28)/3= -0.67

P(Z≤0.67)-P(Z≤-0.67)= 0.749 - 0.251= 0.498

The probability of the tire being inflated within recommended inflation range is 0.498.

I hope this helps!

5 0

2 years ago

Two students from a group of eight boys and 12 girls are sent to represent the school in a parade. If the students are chosen at

kari74 [83]

Answer:

Step-by-step explanation:

* Lets explain how to find the probability of an event

- The probability of an Event = Number of favorable outcomes ÷ Total

number of possible outcomes

- P(A) = n(E) ÷ n(S) , where

# P(A) means finding the probability of an event A

# n(E) means the number of favorable outcomes of an event

# n(S) means set of all possible outcomes of an event

- Probability of event not happened = 1 - P(A)

- P(A and B) = P(A) . P(B)

* Lets solve the problem

- There is a group of students

- There are 8 boys and 12 girls in the group

∴ There are 8 + 12 = 20 students in the group

- The students are sent to represent the school in a parade

- Two students are chosen at random

∴ P(S) = 20

- The students that chosen are not both girls

∴ The probability of not girls = 1 - P(girls)

∵ The were 20 students in the group

∵ The number of girls in the group was 12

∴ The probability of chosen a first girl = 12/20

∵ One girl was chosen, then the number of girls for the second

choice is less by 1 and the total also less by 1

∴ The were 19 students in the group

∵ The number of girls in the group was 11

∴ The probability of chosen a second girl = 11/19

- The probability of both girls is P(1st girle) . P(2nd girl)

∴ The probability of both girls = (12/20) × (11/19) = 33/95

- To find the probability of both not girls is 1 - P(both girls)

∴ P(not both girls) = 1 - (33/95) = 62/95

* The probability that the students chosen are not both girls is 62/95

3 0

2 years ago

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Customers enter the waiting line to pay for food as they leave a cafeteria on a first-come, first-served basis. The arrival rate

Elan Coil [88]

Answer:

The answer is B) 0.57.

Step-by-step explanation:

In this problem we have to apply queueing theory.

It is a single server queueing problem.

The arrival rate is $\lambda=4$ and the service rate is $\mu=7$ .

The proportion of time that the server is busy is now as the "server utilization"and can be calculated as:

$p=\frac{\lambda}{c\mu} =\frac{4}{1*7}=0.57$

where c is the number of server (in this case, one server).

3 0

2 years ago

For which pairs of functions is (f circle g) (x)?

iVinArrow [24]

Answer:

I just took the test and aint gon cap i guessed on it but the answer is B.

f(c)=2/x and g(x)=2/x

sorry I late but i thought i should still tell

Step-by-step explanation:

4 0

2 years ago

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