The superintendent of a large school district, having once had a course in probability and statistics, believes that the number of teachers absent on any given day has a Poisson distribution with parameter µ.
Use the accompanying data on absences for 50 days to obtain a 95% large sample CI for µ.
Absences: 0 1 2 3 4 5 6 7 8 9 10
Frequency: 2 3 8 11 8 7 6 2 1 1 1
[Hint: The mean and variance of a Poisson variable both equal m, so
Z = (X - µ)/√(µ/n),
has approximately a standard normal distribution. Now proceed as in the derivation of the interval for p by making a probability statement (with probability 1- a) and solving the resulting inequalities for µ.

The confidence interval required is (3.37,4.47)

Step-by-step explanation:

From the given data

Let X represents the absence and f is the frequency

Summation of f= 50

Summation of fX=196

Mean of a Poisson distribution= 196/50= 3.92

Critical value

At 5% level of significance for a two tailed z-distribution

Z(0.05/2) = + or - 1.96

The 95% confidence interval sample confidence interval

Cl(95%)= mean+or - z(at 0.05/2)× √(mean/n

Where n=50

Cl(95%)=3.92+/-1.96(√3.92/50)

= (3.37,4.47)

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3