Question

The random variable X, representing the number of cherries in a cherry puff, has the following probability distribution:
X 4 5 6 7
P(X-x)= 0.2 0.4 0.3 0.1
(a) Find the mean of the random variable X.
(b) Find the standard deviation of the random variable X.
(c) Find the probability that the average number of cherries in 36 cherry puffs will not be less than 5.
(d) Find the probability that the difference in the average number of cherries in two randomly selected groups of 30 and 36 cherry puffs, respectively, will be more than 2

Answer 1

Answer:

Explained below.

Step-by-step explanation:

(a)

The mean of the random variable X is:

$\mu=\sum {x\cdot p(x)}=5.3$

(b)

The standard deviation of the random variable X is:

$\sigma=\sqrt{\sum x^{2} p(x)-\mu^{2}}=\sqrt{28.9-(5.3)^{2}}=0.90$

(c)

Compute the probability that the average number of cherries in 36 cherry puffs will not be less than 5 as follows:

Since the sample size is large i.e. n > 30, the sampling distribution of sample mean will follow a normal distribution.

$P(\bar X\leq 5)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}\leq \frac{5-5.3}{0.9/\sqrt{36}})\\\\=P(Z\leq -2)\\\\=1-P(Z<2)\\\\=1-0.97725\\\\=0.02275\\\\\approx 0.023$

(d)

Assume that the two groups are independent.

Then:

$E(D)=5.3-5.3=0$

$SD(D)=\sqrt{\frac{0.9^{2}}{30}+\frac{0.9^{2}}{36}}=0.2225$

Compute the probability that the difference in the average number of cherries will be more than 2 as follows:

$P(D>2)=P(\frac{D-E(D)}{SD(D)}>\frac{2-0}{0.2225})\\\\=P(Z>8.98)\\\\=0$