Question

The length of time to complete a door assembly on an automobile factory assembly line is normally distributed with mean μ=7.3 minutes and standard deviation σ=2 minutes. Samples of size 70 are taken. What is the mean value for the sampling distribution of the sample means?

The length of time to complete a door assembly on an automobile factory assembly line is normally distributed with mean μ=7.2 minutes and standard deviation σ=2.4 minutes. Samples of size 100 are taken. To the nearest thousandth of a minute, what is the standard deviation of the sampling distribution of the sample means?

Find P(46≤x¯≤53) for a random sample of size 35 with a mean of 51 and a standard deviation of 12.   (Round your answer to four decimal places.)

Answer 1

(A)Solution :

Given that,

mean = = 7.3

standard deviation = = 2

n = 70

= 7.3

answer = 7.3

(B)Solution :

Given that,

mean = = 7.2

standard deviation = = 2.4

n = 100

= / n = 24 / 100=2.400

(C)

)Solution :

Given that,

mean = = 51

standard deviation = = 12

n = 35

= / n = 12 / 35=2.0284

P(46< <53 ) = P[(46 - 51) /2.0284 < ( - ) / < (53 - 51) /2.0284 )]

= P(-2.47 < Z <0.99 )

= P(Z <0.99 ) - P(Z < -2.47)

Using z table,  

= 0.8389 -0.0068   

= 0.8321