7. No Broken Hearts [2.5 Points]Robert has a standard deck of 52 playing cards. Robert shuffles the deck, draws a card, andcheck
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Answer:
(a) The mean and standard deviation of <em>X</em> is 2.6 and 1.2 respectively.
(b) The mean and standard deviation of <em>T</em> are 390 and 180 respectively.
(c) The distribution of <em>T</em> is <em>N</em> (390, 180²). The probability that all students’ requests can be accommodated is 0.7291.
Step-by-step explanation:
(a)
The random variable <em>X</em> is defined as the number of tickets requested by a randomly selected graduating student.
The probability distribution of the number of tickets wanted by the students for the graduation ceremony is as follows:
X P (X)
0 0.05
1 0.15
2 0.25
3 0.25
4 0.30
The formula to compute the mean is:
Compute the mean number of tickets requested by a student as follows:
The formula of standard deviation of the number of tickets requested by a student as follows:
Compute the standard deviation as follows:
Thus, the mean and standard deviation of <em>X</em> is 2.6 and 1.2 respectively.
(b)
The random variable <em>T</em> is defined as the total number of tickets requested by the 150 students graduating this year.
That is, <em>T</em> = 150 <em>X</em>
Compute the mean of <em>T</em> as follows:
Compute the standard deviation of <em>T</em> as follows:
Thus, the mean and standard deviation of <em>T</em> are 390 and 180 respectively.
(c)
The maximum number of seats at the gym is, 500.
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.
Here <em>T</em> = total number of seats requested.
Then, the mean of the distribution of the sum of values of X is given by,
And the standard deviation of the distribution of the sum of values of X is given by,
So, the distribution of <em>T</em> is N (390, 180²).
Compute the probability that all students’ requests can be accommodated, i.e. less than 500 seats were requested as follows:
Thus, the probability that all students’ requests can be accommodated is 0.7291.