Question

The random variable X, representing the number of cherries in a cherry puff, has the following probability distribution:

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.

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

(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:

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