Answer:
Step-by-step explanation:
From the information given,
Number of personnel sampled, n = 85
Mean or average = 6.5
Standard deviation of the sample = 1.7
We want to determine the confidence interval for the mean number of years that personnel spent in a particular job before being promoted.
For a 95% confidence interval, the confidence level is 1.96. This is the z value and it is determined from the normal distribution table. We will apply the following formula to determine the confidence interval.
z×standard deviation/√n
= 1.96 × 6.5/√85
= 1.38
The confidence interval for the mean number of years spent before promotion is
The lower end of the interval is 6.5 - 1.38 = 5.12 years
The upper end is 6.5 + 1.38 = 7.88 years
Therefore, with 95% confidence interval, the mean number of years spent before being promoted is between 5.12 years and 7.88 years
Answer:
(x + 3, y - 2)
Step-by-step explanation:
we know that
The rule of the transformation T is equal to
T: pre-image → image
T: (x, y) → (x',y')
T: (x, y) → (x - 3, y + 2)
so
x'=x-3 ----> x=x'+3
y'=y+2 ----> y=y'-2
The rule of the inverse of transformation T -1 is equal to
T-1: image → pre-image
T-1: (x', y') → (x,y)
T-1: (x', y') → (x'+3, y'-2)
Let
L=event that a selected worker has low risk
H=event that a selected worker has high risk
We need to find
P(HL)+P(LH)
=5*20/(25*24)+20*5/(25*24)
=1/6+1/6
=1/3