Answer:
Step-by-step explanation:
<pGreetings!a. The variable X represents the height of a Goomba, which follows a normal distribution with a mean of μ= 12 inches and a standard deviation of δ= 6 inches.
To find the probability that a Goomba picked at random has a height between 13 and 15 inches, you express it as:
P(13≤X≤15)
Considering that standard normal probability tables provide cumulative values, you can express this range as the cumulative probability up to 15 minus the cumulative probability up to 13. You'll first need to standardize these variable heights to obtain corresponding Z values:
P(X≤15) - P(X≤13)
P(Z≤(15-12)/6) - P(Z≤(13-12)/6)
P(Z≤0.33) - P(Z≤0.17)= 0.62930 - 0.56749= 0.06181
b. Now we have Y as the variable indicating the height of a Koopa Troopa. This variable also follows a normal distribution, with a mean μ= 15 inches and a standard deviation δ=3 inches.
The query concerns the probability that a Koopa Troopa stands taller than 75% of Goombas.
First step:
You need to determine the height of a randomly chosen Koopa Troopa that exceeds 75% of the Goomba population.
This entails determining the value of X corresponding to the limit below which 75% of the population falls, denoted by:
P(X ≤ b)= 0.75
Step 2:
Search the standard normal distribution for the Z value that has 0.75 beneath it:
Next, you will reverse the standardization to solve for "b"
Z= (b - μ)/δ
b= (Z*δ)+μ
b= (0.674*6)+12
b= 16.044 inches
Step 3:
With the height that identifies a Koopa Troopa taller than 75% of the Goomba population determined, compute the probability of selecting that Koopa Troopa:
P(Y≤16.044)
This time, utilize the Koopa’s average height and standard deviation to find the probability:
P(Z≤(16.044-15)/3)
P(Z≤0.348)= 0.636
The likelihood of randomly selecting a Koopa Troopa that is taller than 75% of Goombas is 63.6%
I hope this information is useful!
Answer:
840
Step-by-step explanation:
As the arrangement matters, we apply the permutations formula to find the solution.
Permutations formula:
The count of possible arrangements of x items chosen from a total of n items is defined by this formula:
For this problem:
Jose occupies the first seat.
The other four can be arranged among the remaining 7 family members. Thus
Hence, the final answer is:
840
Response:
a. As student debt rises, current investment diminishes.
b. Y= 68778.2406 - 1.9112X
For each dollar increase in college debt, the average current investments decrease by 1.9112 dollars.
c. A substantial linear correlation exists between college debt and current investment as the P-value falls below 0.1.
d. Y= $59222.2406
e. R²= 0.9818
Step-by-step breakdown:
Hello!
Data has been gathered on a random sample of 20 individuals who completed their college education five years ago. The variables under consideration are:
Y: Current investment by an individual who graduated from college five years prior.
X: Total debt of an individual upon graduating five years ago.
a)
To explore the relationship between debt and investment, creating a scatterplot with the sample data is ideal.
The scatterplot demonstrates a negative correlation, indicating that as these individuals' debt increases, their current investments decrease.
Therefore, the statement that accurately describes this is: As college debt rises, current investment decreases.
b)
The population regression equation is Y= α + βX +Ei
To develop this equation, estimates for alpha and beta are required:
a= Y[bar] -bX[bar]
a= 44248.55 - (-1.91)*12829.70
a= 68778.2406
b=
b=
b= -1.9112
∑X= 256594
∑X²= 4515520748
∑Y= 884971
∑Y²= 43710429303
∑XY= 9014653088
n= 20
Averages:
Y[bar]= ∑Y/n= 884971/20= 44248.55
X[bar]= ∑X/n= 256594/20= 12829.70
The estimated regression equation becomes:
Y= 68778.2406 - 1.9112X
For every dollar increase in college debt, the average current investments drop by 1.9112 dollars.
c)
To evaluate if there's a linear regression between these variables, the following null hypotheses are formulated:
H₀: β = 0
H₁: β ≠ 0
α: 0.01
Testing can be performed utilizing either a Student t-test or Snedecor's F (ANOVA)
Using t= b - β = -1.91 - 0 = -31.83
Sb 0.06
The critical area and P-value for this test is two-tailed. The P-value equals: 0.0001
Since this P-value is underneath the significance level, we reject the null hypothesis.
In the case of ANOVA, the rejection area is also one-tailed to the right, corresponding to the P-value.
The P-value remains: 0.0001
Using this method, we similarly reject the null hypothesis.
This indicates that 98.18% of the variability in current investments relates to college graduation debt within the projected regression model: Y= 68778.2406 - 1.9112X
I trust this is beneficial!
