Answer:
Number of student tickets sold are 57 and general tickets sold are 68.
Step-by-step explanation:
Let the number of student tickets = x and the number of general tickets = y.
It is given that in total 125 tickets were sold.
So, we have,
Also, students tickets is sold for $6 each and the general tickets are sold for $10 each.
Since, the total cost of the tickets is $1022.
So, we get,
Thus, the system of equations obtained is,
x + y = 125 ........................(1)
6x + 10y = 1022
Multiply equation (1) by 6 gives us,
6x + 6y = 750 ............................(2)
6x + 10y = 1022 .........................(3)
Subtact (2) from (3), we get,
10y - 6y = 1022 - 750
i.e. 4y = 272
i.e. y= 68.
So, equation (1) gives,
x = 125 - y
i.e. y = 125 - 68
i.e. y = 57
Hence, the number of student tickets sold are 57 and general tickets sold are 68.
Answer:
Mean = 3.12, Variance = 3.324, Standard deviation = 1.8232
Step-by-step explanation:
Total number of students = 10 students.
Given data, 1.1, 5.2, 3.6, 5.0, 4.8, 1.8, 2.2, 5.2, 1.5, 0.8
To find the mean, at first we have to take the sum of all given data and then divide with the number of students.
Let the data is X, = 1.1, + 5.2, + 3.6, + 5.0, + 4.8, + 1.8, + 2.2, + 5.2, + 1.5, + 0.8 = 31.2
Mean = 31.2 / 10 = 3.12
Answer:
58.32% probability that a randomly selected application will report a GMAT score of less than 600
93.51% probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600
98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
What is the probability that a randomly selected application will report a GMAT score of less than 600?
This is the pvalue of Z when X = 600. So
has a pvalue of 0.5832
58.32% probability that a randomly selected application will report a GMAT score of less than 600
What is the probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600?
Now we have
This is the pvalue of Z when X = 600. So
has a pvalue of 0.9351
93.51% probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600
What is the probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600?
Now we have
has a pvalue of 0.9838
98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600