Answer:
a) 0.00019923%
b) 47.28%
Step-by-step explanation:
a) To find the probability of all sockets in the sample being defective, we can do the following:
The first socket will be in a group where 5 of the 38 sockets are defective, so the probability is 5/38
The second socket will be in a group where 4 of the 37 sockets are defective, as the first one picked is already defective, so the probability is 4/37
Expanding this, we have that the probability of having all 5 sockets defective is: (5/38)*(4/37)*(3/36)*(2/35)*(1/34) = 0.0000019923 = 0.00019923%
b) Following the same logic of (a), the first socket have a chance of 33/38 of not being defective, as we will pick it from a group where 33 of the 38 sockets are not defective. The second socket will have a chance of 32/37, and so on.
The probability will be (33/38)*(32/37)*(31/36)*(30/35)*(29/34) = 0.4728 = 47.28%
Oooooooonnnnnneeeeeeeemmmmmmmmiiiiilllllllliiiiieeee
Answer:
a) 5.5
b) None
Step-by-step explanation:
The given data set is {96,89,79,85,87,94,96,98}
First we must find the mean.

We now find the absolute value of the distance of each value from the mean.
This is called the absolute deviation
{
}
{
}
We now find the mean of the absolute deviations

The least absolute deviation is 1.5. This is not within one absolute deviation.
Therefore none of the data set is closer than one mean absolute deviation away from the mean.
Answer:
The number of bacteria
after
days is given by

where
is the initial number of bacteria.
Step-by-step explanation:
The number of bacteria
in the sample triples every 10 days, this means after the first 10th day, the number of bacteria is

where
is the initial number of bacteria in the sample.
After the 2nd 10th days, the number of bacteria is

after the 3rd day,

and so on.
Thus, the formula we get for the number of bacteria after the <em>n</em>th 10-days is

where
is is the <em>n</em>th 10-days.
Since,
is 10 days, we have

or

Substituting that into
, we get:
