Question

your statistics professor hands you a fair die, with each side having the same chance of being the uppermost face. you are asked to toss the die 7 times and count the number of times the die shows a topside of six.
this count is represented by the random variable (x).
(a) complete the probability - distribution table of (x) below. use four decimals in each of your entries.

(x)	0	1	2	3	4	5	6	7
(p(x = x))							0.00000357

(b) from the distribution you found in part (a), what can you say about the distribution of (x)?
the distribution of (x) is? with an expected value of sixes and a standard deviation of sixes.
(enter your answers to two decimals.)
(c) as requested, you tossed the die 7 times and observed (x = 6) sixes. if you were to repeat the 7 tosses of this die, what is the probability that you will observe at least 6 sixes again? enter your answer to four decimal places.
(d) in part (a) you gave the probability distribution for how many times a fair die tossed 7 times should show a six. lets say you tried this experiment, and 6 out of the 7 rolls landed on a six (i.e. (x = 6)). feeling suspicious, you try another 7 times... and again got either 6 or all 7 (out of 7) rolls to land on a six (i.e. (xgeq6)). compare with what should likely have happened (refer to part (a)). what would you say this information means? select the most appropriate answer.
a. the die appears to favor an outcome of a six, since (x = 6) should be an unusual event, yet i continue to observe it.
b. the distribution of the random variable (x) appears to be skewed to the right.
c. the die appears not to be fair. (x = 6) is a usual event, which explains why im seeing it so often.
d. the distribution of the random variable (x) appears to be roughly symmetrical.

Explanation:

Step1: Identify the distribution type

The problem involves a binomial distribution since we have a fixed - number of independent trials ($n = 7$), each with two possible outcomes (rolling a six or not rolling a six), and a constant probability of success ($p=\frac{1}{6}$). The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Calculate probabilities for part (a)

For $k = 0$:
$C(7,0)=\frac{7!}{0!(7 - 0)!}=1$, $p=\frac{1}{6}$, $1 - p=\frac{5}{6}$, $n = 7$.
$P(X = 0)=C(7,0)\times(\frac{1}{6})^{0}\times(\frac{5}{6})^{7}=1\times1\times(\frac{5}{6})^{7}\approx0.2791$.
For $k = 1$:
$C(7,1)=\frac{7!}{1!(7 - 1)!}=7$, $P(X = 1)=C(7,1)\times(\frac{1}{6})^{1}\times(\frac{5}{6})^{6}=7\times\frac{1}{6}\times(\frac{5}{6})^{6}\approx0.3609$.
For $k = 2$:
$C(7,2)=\frac{7!}{2!(7 - 2)!}=21$, $P(X = 2)=C(7,2)\times(\frac{1}{6})^{2}\times(\frac{5}{6})^{5}=21\times\frac{1}{36}\times(\frac{5}{6})^{5}\approx0.2344$.
For $k = 3$:
$C(7,3)=\frac{7!}{3!(7 - 3)!}=35$, $P(X = 3)=C(7,3)\times(\frac{1}{6})^{3}\times(\frac{5}{6})^{4}=35\times\frac{1}{216}\times(\frac{5}{6})^{4}\approx0.0918$.
For $k = 4$:
$C(7,4)=\frac{7!}{4!(7 - 4)!}=35$, $P(X = 4)=C(7,4)\times(\frac{1}{6})^{4}\times(\frac{5}{6})^{3}=35\times\frac{1}{1296}\times(\frac{5}{6})^{3}\approx0.0195$.
For $k = 5$:
$C(7,5)=\frac{7!}{5!(7 - 5)!}=21$, $P(X = 5)=C(7,5)\times(\frac{1}{6})^{5}\times(\frac{5}{6})^{2}=21\times\frac{1}{7776}\times(\frac{5}{6})^{2}\approx0.0026$.
For $k = 6$:
$C(7,6)=\frac{7!}{6!(7 - 6)!}=7$, $P(X = 6)=C(7,6)\times(\frac{1}{6})^{6}\times(\frac{5}{6})^{1}=7\times\frac{1}{46656}\times\frac{5}{6}\approx0.0001$.
For $k = 7$:
$C(7,7)=\frac{7!}{7!(7 - 7)!}=1$, $P(X = 7)=C(7,7)\times(\frac{1}{6})^{7}\times(\frac{5}{6})^{0}=(\frac{1}{6})^{7}\approx0.00000357$.

Step3: Calculate expected value and standard deviation for part (b)

The expected value of a binomial distribution $E(X)=np$, so $E(X)=7\times\frac{1}{6}\approx1.17$.
The standard deviation $\sigma=\sqrt{np(1 - p)}=\sqrt{7\times\frac{1}{6}\times\frac{5}{6}}\approx0.98$.
The distribution of $X$ is binomial.

Step4: Calculate probability for part (c)

$P(X\geq6)=P(X = 6)+P(X = 7)=0.0001 + 0.00000357\approx0.0001$.

Step5: Analyze the result for part (d)

The probability of $X = 6$ is very low, but it is being observed repeatedly. So the die appears to favor an outcome of a six, since $X = 6$ should be an unusual event, yet it is being observed.

Answer:

(a)

$x$	0	1	2	3	4	5	6	7

(b) binomial, 1.17, 0.98
(c) 0.0001
(d) A. The die appears to favor an outcome of a six, since $X = 6$ should be an unusual event, yet I continue to observe it.