Question

The length of time to complete a door assembly on an automobile factory assembly line is normally distributed with mean μ = 6.7 minutes and standard deviation σ = 2.2 minutes.

What is the probability that one door takes less than 6 minutes to assemble?

A sample of 2000 is taken, what is the mean value for this sampling distribution of sample means?

A sample of size 400 is taken, what is the standard error of this sampling distribution of sample means?

A sample of size 20 is taken, what is the shape of the sampling distribution of sample means?

A sample of 400 is taken, what is the probability that the sample mean is between 6.5 and 7.2 minutes?

A sample of 400 is taken, what is the probability that the sample mean is less than 6.4?

Answer 1

let, X: length of time to complete door assembly XON(ll= 6-7 , O = 2.2) = 0.37526 0.3752 B A) P(X<6) = P(Z < 6-6-7 P(Z < 6-6.7) = P(Z2-0.3182 <(foom t table PCXCO) If n=2000 then mean value will remain same for the Sampling distigbretiona i.e. llx = 6.7 c) It Standard etzat 10.11 to nz400 2.2 1400 a) of n=20 the shape of & distribution of mean will be noemal. Creo bell shaped) to Centeal limit theorem) sampling ( According

e) = 400 Then 1 7.2-6.7 we need to find P(6.5<x<7.2) = P(4,5.5,6,7 424 2077400 P(-1.8182 << <4.5454 202/3400 = P(Z24.5454 - PlZX-108182 1 - 0.0345 → (from z-table) = 0.9655 n --400 Then we need to P(X< 6.4) find, PC Z < 6.4- 6.7 262/7400 Plze=2.7273 = 0.0082
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