91影视

It is given that 16% of the M&M candies are green, as claimed by the manufacturer.

A set of 100 candies is provided, out of which 19 are green.

The total number of candies present (n) is equal to 100.

The probability of selecting a green candy is given to be equal to

$$\begin{gathered} p=16\% \\ =\frac{16}{100} \\ =0.16 \end{gathered}$$

The mean of the number of green candies is calculated as follows:

$$\begin{gathered} 渭=np \\ =100\left(0.16\right) \\ =16 \end{gathered}$$

Thus, $渭=16$.

The standard deviation for the number of green candies is computed below:

$$\begin{gathered} 蟽=\sqrt{npq} \\ =\sqrt{np\left(1-p\right)} \\ =\sqrt{100脳0.16脳\left(1-0.16\right)} \\ =3.66 \end{gathered}$$

Thus, $蟽=3.66$.

a.

The following limit separates significantly low values:

$$\begin{gathered} 渭-2蟽=16-\left(2\right)\left(3.66\right) \\ =8.68 \\ 鈮�8.7 \end{gathered}$$

Therefore, values less than or equal to 8.7 are considered to be significantly low.

The following limit separates significantly high values:

$$\begin{gathered} 渭+2蟽=16+\left(2\right)\left(3.66\right) \\ =23.32 \\ 鈮�23.3 \end{gathered}$$

Therefore, values greater than or equal to 23.3 are considered to be significantly high.

Values lying between 8.67 and 23.33 are not significant.

Here, the value of 19 lies between 8.7 and 23.3. Thus, it cannot be considered significantly high.

b.

Let X denote the number of green candies.

Success is defined as selecting a green candy.

The probability of success is p=0.16.

The probability of failure is computed below:

$$\begin{gathered} q=1-p \\ =1-0.16 \\ =0.84 \end{gathered}$$

The number of trials (n) is equal to 100.

The binomial probability formula used to compute the given probability is as follows:

$P\left(X=x\right)={}^n{C}_x{\left(p\right)}^x{\left(q\right)}^{n-x}$

By using the binomial probability formula, the probability of selecting 19 green candies can be calculated in the following manner:

$$\begin{gathered} P\left(X=19\right)={}^{100}{C}_{19}{\left(0.16\right)}^{19}{\left(0.84\right)}^{100-19} \\ =0.0736 \end{gathered}$$

Thus, the probability of selecting 19 green candies is equal to 0.0736.

c.

The probability of 19 or more green candies has the following expression:

$$\begin{gathered} P\left(X猢�19\right)=1-P\left(X<19\right) \\ =1-\left[P\left(X=0\right)+P\left(X=1\right)+.......+P\left(X=18\right)\right] \end{gathered}$$

The individual probabilities will be computed as follows:

$$\begin{gathered} P\left(X=0\right)={}^{100}{C}_0{\left(0.16\right)}^0{\left(0.84\right)}^{100} \\ 鈮�0.00000003 \\ P\left(X=1\right)={}^{100}{C}_1{\left(0.16\right)}^0{\left(0.84\right)}^{100} \\ 鈮�0.00000051 \\ . \\ . \\ . \\ P\left(X=18\right)={}^{100}{C}_{18}{\left(0.16\right)}^{18}{\left(0.84\right)}^{82} \\ 鈮�0.089474068 \end{gathered}$$

Thus, the required probability is computed as follows:

$$\begin{gathered} P\left(X猢�19\right)=1-P\left(X<19\right) \\ =1-\left[P\left(X=0\right)+P\left(X=1\right)+.......+P\left(X=18\right)\right] \\ =1-0.757559073 \\ =0.242 \end{gathered}$$

Thus, the probability of selecting 19 or more green candies is equal to 0.242.

d.

The probability formula to determine whether the given value of the number of successes (x) is significantly high or not is shown below:

$P\left(x\text{or}\text{more}\right)猢�0.05$

Here, the considered value of x is equal to 19.

Since the probability of 19 or more green candies is represented in part (c), it can be concluded that theprobability computed in part (c) is relevant for determining whether the result of 19 green M&Ms is significantly high.

Thus,

$$\begin{gathered} P\left(19\text{or}\text{more}\right)=0.242 \\ >0.05 \end{gathered}$$

Since the probability of 19 or more green candies is not less than or equal to 0.05, the value of 19 green candies is not significantly high.

e.

As the results for the significance of 19 green candies are in accordance with the limits computed using p=0.16, it can be said that the results do not provide enough evidence against the claim of 16% for green candies.