If 25 servings of guacamole have been sold from a batch, in ounces, what is the standard deviation of how much is left? (Suppose the servings of guacamole are independent.)

The standard deviation of the the left guacamole is 12.51.

Step-by-step explanation:

As the complete question is not given, thus the complete question is found online and is attached herewith.

From the data

Population mean==2.2 ounces

Population Standard Deviation==0.4 ounces

Batch mean==80 ounces

Batch Standard Deviation==7 ounces

As per serving is 2.2 ounces

So for 25 servings, Guacamole used is

25*2.2=55 ounces

Thus the Guacamole left is given as

Batch-55=80-55-25 ounces

As it is given that the servings are independent, thus

Var(Z)=Var(X)+Var(Y)

Here Var(Z) is the variance of the Batch, X is the served guacamole and Y is the left guacamole

Also as variance is given as so putting the values

So the standard deviation of the the left guacamole is 12.51.

0 and 5 minutes

step-by-step explanation:

given that one morning eight buses arrive late by the number of minutes given below:

0 8 2 6 9 2 0 7 8

write these in ascending order

0   0 2 6 7 8 8 9

in the afternoon two more buses arrive at the bus stop say by x and y minutes late.

total number of buses = 10

median = average of 4th and 5th after arranging in ascending order

since 4th entry is 6, the values of x and y should be below 6

also given 0 is the mode so either x or y or both should be 0 because then only 0 gets the maximum frequency and hence mode

thus we have

0 0 0 2 6 7 8 8 9 and another unknown entry which is below 6.

since median is 3.5, we have the unknown entry to be 5 only because then only median =( 2+5)/2

thus the after noon buses were late by 0 and 5 minutes.