Plis tell me if it’s correct and do I need to divide by 2 at the end like I did?

The amounts a soft drink machine is designed to dispense for each drink are normally distributed, with a mean of 12.1 fluid ounc

Ludmilka [50]

Answer:

$\mu = 12.1\\\sigma = 0.3$

a) Find the probability that the drink is less than 11.9 fluid ounces

We are supposed to find P(x<11.9)

x= 11.9

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{11.9-12.1}{0.3}$

$z=-0.67$

Refer the z table

So, P(z<-0.67) = 0.2514

Hence the probability that the drink is less than 11.9 fluid ounces is 0.2514

b)Find the probability that the drink is between 11.6 and 11.9 fluid ounces

x= 11.6

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{11.6-12.1}{0.3}$

$z=-1.67$

x= 11.9

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{11.9-12.1}{0.3}$

$z=-0.67$

We are supposed to find P(11.6<x<11.9)

So, P(11.6<x<11.9)= P(x<11.9)-P(x<11.6)

P(-1.67<z<-0.67)= P(z<-0.67)-P(z<-1.67)

Refer the z table

P(-1.67<z<-0.67)= P(z<-0.67)-P(z<-1.67)

P(-1.67<z<-0.67)= 0.2514-0.0475

P(-1.67<z<-0.67)= 0.2039

So, the probability that the drink is between 11.6 and 11.9 fluid ounces is 0.2039.

c)Find the probability that the drink is more than 12.6 fluid ounces

We are supposed to find P(x>12.6)

x= 12.6

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{12.6-12.1}{0.3}$

$z=1.67$

Refer the z table

P(x>12.6)=1-P(x<12.6)

P(z>1.67)=1-P(z<1.67)

P(z>1.67)=1-0.9525

P(z>1.67)=0.0475

Hence the probability that the drink is more than 12.6 fluid ounces is 0.0475

d) Is the drink containing more than 12.6 fluid ounces an unusual event?

Since p value is less than alpha (0.05)

Answer = No, because the probability that a drink containing more than 12.6 fluid ounces is less than 0.05, this event is not unusual.

3 0

3 years ago

Suppose three marksmen shoot at a target. The ith marksman fires ni times, hitting the target each time with probability Pi, ind

astraxan [27]

Answer:

Distribution is NOT binomial.

Step-by-step explanation:

In order to be a binomial distribution, the probability of success for each individual trial must be the same. Since each marksman hits the target with probability Pi, the probability of success (hitting the target) is not necessarily equal for all trials. Therefore, the distribution is not binomial.

In this case, the distribution would only be binomial if Pi was the same for every "ith" marksman.

6 0

3 years ago

Find the equation for a parabola if the vertex is 6,9

devlian [24]

The equation for a parabola with vertex (h, k) and vertical scale factor "a" is
y = a(x -h)² + k

One parabola with vertex (6, 9) is
y = (x-6)² +9