The stem-and-leaf plot below shows the number of pages each student in the class read the previous evening.
2 answers:
The stem-leaf plot is attached below
Answer:
According to the plot, 75% of students read less than 35 pages ⇒ 3rd answer
Step-by-step explanation:
At first lets count the number of the students from the picture
There are 4 students in the 1st row
There are 6 students in the 2nd row
There are 6 students in the 3rd row
There are 3 students in the 4th row
There are 3 students in the 5th row
There are 1 student in the 6th row
The total number of students = 4 + 6 + 6 + 3 + 3 + 1 = 23
Let us find 75% of them equals how many students
75% × 23 = × 23 = 17.25
That means 75% of the students is about 17 students
Let us find the 17th student read how many pages
From the chart the 17th students read 33 pages
The 18th student read 35 pages
So 75% of the students read less than 35 pager
According to the plot, 75% of students read less than 35 pages
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Answer:
a) the probability that the defective board was produced during the first hour of operation is or 0.1000
b) the probability that the defective board was produced during the last hour of operation is or 0.1000
c) the required probability is 0.2000
Step-by-step explanation:
Given the data in the question;
During a specific ten-hour period, one defective circuit board was found.
Lets X represent the number of defective circuit boards coming out of the machine , following Poisson distribution on a particular 10-hours workday which one defective board was found.
Also let Y represent the event of producing one defective circuit board, Y is uniformly distributed over ( 0, 10 ) intervals.
f(y) = ; ( a ≤ y ≤ b )
= ; ( 0 ≤ y ≤ 10 )
f(y) = ; ( 0 ≤ y ≤ 10 )
Now,
a) the probability that it was produced during the first hour of operation during that period;
P( Y < 1 ) =
we substitute
=
=
=
= or 0.1000
Therefore, the probability that the defective board was produced during the first hour of operation is or 0.1000
b) The probability that it was produced during the last hour of operation during that period.
P( Y > 9 ) =
we substitute
=
=
=
= or 0.1000
Therefore, the probability that the defective board was produced during the last hour of operation is or 0.1000
c)
no defective circuit boards were produced during the first five hours of operation.
probability that the defective board was manufactured during the sixth hour will be;
P( 5 < Y < 6 | Y > 5 ) = P[ ( 5 < Y < 6 ) ∩ ( Y > 5 ) ] / P( Y > 5 )
= P( 5 < Y < 6 ) / P( Y > 5 )
we substitute
= ( 6-5 ) / ( 10 - 5 )
= 0.2000
Therefore, the required probability is 0.2000