Question

A population of values has a normal distribution with μ=102.8μ=102.8 and σ=41.9σ=41.9. You intend to draw a random sample of size n=88n=88.

Find the probability that a single randomly selected value is greater than 100.6.
P(X > 100.6) =

Find the probability that a sample of size n=88n=88 is randomly selected with a mean greater than 100.6.
P(M > 100.6) =

Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 204.1-cm and a standard deviation of 1.1-cm. For shipment, 27 steel rods are bundled together.

Find the probability that the mean length of a randomly selected bundle of steel rods is greater than 203.6-cm.
P(¯xx¯ > 203.6-cm) =

Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 2 decimal places are accepted.

Answer 1

Solution :

1 ) Given that,

mean = = 102.8

standard deviation = = 41.9

a ) P (x > 100.6)

= 1 - P (x < 100.6 )

= 1 - P ( x -  / ) < ( 100.6 - 102.8 / 41.9)

= 1 - P ( z < -2.2 / 41.9)

= 1 - P ( z < -0.052)

Using z table

= 1 - 0.4793

= 0.5207

Probability = 0.5207

b ) n = 88

= 1200

₌ / n = 88 41.9 = 13.5949

P (M > 100.6)

= 1 - P (M < 100.6 )

= 1 - P ( M - / ) <-0.162 ( 100.6 - 102.8 / 13.5949 )

= 1 - P ( z < -2.2 / 13.5949 )

= 1 - P ( z < -0.162 )

Using z table

= 1 - 0.4793

= 0.5643

Probability = 0.5643

2 ) Given that,

mean = = 204.1

standard deviation = = 1.1

P (x > 203.6)

= 1 - P (x < 203.6 )

= 1 - P ( x -  / ) < ( 203.6 - 204.1 / 1.1 )

= 1 - P ( -0.5 / 1.1)

= 1 - P ( z < -0.454)

Using z table

= 1 - 0.3249

= 0.6751

Probability = 0.6751