A figure is made up of a square and a rectangle. The area of the square is 64 cm². The breadth of the rectangle is the same leng
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Solution :
Given :
Sample mean,
Sample size, n = 129
Sample standard deviation, s = 8.2
a. Since the population standard deviation is unknown, therefore, we use the t-distribution.
b. Now for 95% confidence level,
α = 0.05, α/2 = 0.025
From the t tables, T.INV.2T(α, degree of freedom), we find the t value as
t =T.INV.2T(0.05, 128) = 2.34
Taking the positive value of t, we get
Confidence interval is ,
(32.52, 35.8)
95% confidence interval is (32.52, 35.8)
So with confidence of the population of the mean number of the pounds per person per week is between 32.52 pounds and 35.8 pounds.
c. About of confidence intervals which contains the true population of mean number of the pounds of the trash that is generated per person per week and about
that doe not contain the true population of mean number of the pounds of trashes generated by per person per week.
Answer:
a) P(X>np+3=0.017
b) P(x>1)=0.190
c) P(Y>1)=0.651
Step-by-step explanation:
This a binomial experiment where success is denoted by parts that need rework.
X ∼ B(n, p); n = 20; p = 0.01
The expected value of X is: E(X) = np =20×0.01= 0.2
The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,
The standard deviation SD(X)= ≈ 0.445
a) P(X>np+3=P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)
Probability function is given by:
P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - -
P(X≥2)=1-0.165-0.818=0.017
b) p=0.04
P(x>1)=P(x≥2)= 1 - P(x=1) - P(x=0)= 1 - -
P(x>1)= 1 - 0.368 - 0.442=0.190
c) In this case we consider p=0.19 (Probability that X exceeds 1)
In this experiment Y is the number of hours and n= 5 hours.
Then, we check the probability in each hour:
P(Y>1)=1- P(Y=0)
P(Y=0)==0.349
P(Y>1)=1-0.349=0.651