Lalo has 1,500 minutes per month on his cell phone plan. How many more minutes can he use if he has already talked for 785 minut
1 answer:
Answer:
715 minutes
Step-by-step explanation:
we know that
To find out how many more minutes he can use, subtract the number of minutes he has already talked from the number of minutes of his plan
Let
x -----> the number of minutes of his plan
y -----> he number of minutes he has already talked
so
we have
substitute
You might be interested in
Answer:
75%
88.89%
Step-by-step explanation:
Given :
Mean = 70
Standard deviation = 12
Since the data is said to be extremely skewed, we apply Chebyshev's theorem rather than the empirical rule :
The minimum proportion of observation between 46 and 94
Chebyshev's theorem :
1 - 1 / k²
k = number of standard deviations from the mean
k = (94 - 70) / 12 = 24 / 12 = 2
Hence, we have ;
1 - 1/2²
1 - 1/4
1 - 0.25 = 0.75
Hence, The minimum proportion of observation between 46 and 94 is 75%
Between 36 and 106 :
k = (106 - 70) / 12 ;
k = 36/12 = 3
Hence,
1 - 1/3² = 1 - 1/9 = 8/9 = 0.8888 = 88.89%
The minimum proportion of observation between 34 and 106 is 88.89%
Answer:
Step-by-step explanation:
Hello!
For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.
In this example the variable is:
X: height of a college student. (cm)
There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.
The option you have is to apply the Central Limit Theorem.
The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:
X[bar]~~N(μ;σ2/n)
Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:
98% CI
1 - α: 0.98
⇒α: 0.02
α/2: 0.01
X[bar] ±
174.5 ±
[172.22; 176.78]
With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].
I hope it helps!