Mike ran 1/2 mile during gym class . how many feet did he run.

If XX is a binomial random variable, compute the mean, the standard deviation, and the variance for each of the following cases:

Liula [17]

Answer:

(a) The mean, variance and standard deviation of X are 1.60, 0.96 and 0.98 respectively.

(b) The mean, variance and standard deviation of X are 3.20, 0.64 and 0.80 respectively.

(c) The mean, variance and standard deviation of X are 1.50, 0.75 and 0.87 respectively.

(d) The mean, variance and standard deviation of X are 4.00, 0.80 and 0.89 respectively.

Step-by-step explanation:

The random variable X follows a Binomial distribution with parameters n and p.

The mean, variance and standard deviation of X are:

$\mu=np\\\sigma^{2}=np(1-p)\\\sigma=\sqrt{np(1-p)}$

(a)

For n = 4 and p = 0.40 compute the mean, variance and standard deviation of X as follows:

$\mu=np=4\times0.40=1.60\\\sigma^{2}=np(1-p)=4\times0.40\times(1-0.40)=0.96\\\sigma=\sqrt{np(1-p)}=\sqrt{0.96}=0.98$

Thus, the mean, variance and standard deviation of X are 1.60, 0.96 and 0.98 respectively.

(b)

For n = 4 and p = 0.80 compute the mean, variance and standard deviation of X as follows:

$\mu=np=4\times0.80=3.20\\\sigma^{2}=np(1-p)=4\times0.80\times(1-0.80)=0.64\\\sigma=\sqrt{np(1-p)}=\sqrt{0.64}=0.80$

Thus, the mean, variance and standard deviation of X are 3.20, 0.64 and 0.80 respectively.

(c)

For n = 3 and p = 0.50 compute the mean, variance and standard deviation of X as follows:

$\mu=np=3\times0.50=1.50\\\sigma^{2}=np(1-p)=3\times0.50\times(1-0.50)=0.75\\\sigma=\sqrt{np(1-p)}=\sqrt{0.75}=0.87$

Thus, the mean, variance and standard deviation of X are 1.50, 0.75 and 0.87 respectively.

(d)

For n = 5 and p = 0.80 compute the mean, variance and standard deviation of X as follows:

$\mu=np=5\times0.80=4.00\\\sigma^{2}=np(1-p)=5\times0.80\times(1-0.80)=0.80\\\sigma=\sqrt{np(1-p)}=\sqrt{0.80}=0.89$

Thus, the mean, variance and standard deviation of X are 4.00, 0.80 and 0.89 respectively.

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