Question

A population of values has a normal distribution with μ=62.7μ=62.7 and σ=66.2σ=66.2. You intend to draw a random sample of size n=42n=42.

Find the probability that a single randomly selected value is greater than 49.4.
P(X > 49.4) =

Find the probability that a sample of size n=42n=42 is randomly selected with a mean greater than 49.4.
P(M > 49.4) =

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.

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Answer 1

Solution :

Given that,

mean = = 62.7

standard deviation = = 66.2

a )P (x > 49.4 )

= 1 - P (x <49.4 )

= 1 - P ( x -  / ) < ( 49.4 - 62.7 / 66.2)

= 1 - P ( z < -13.3 / 66.2 )

= 1 - P ( z < -0.201 )

Using z table

= 1 - 0.4203

= 0.5797

Probability = 0.5797

b ) n = 42

=62.7

₌ / n = 66.2 42 = 10.2149

P (x > 49.4 )

= 1 - P ( <49.4 )

= 1 - P ( - / ) < ( 49.4 - 62.7 / 10.2149)

= 1 - P ( z < -13.3 / 10.2149 )

= 1 - P ( z <-1.302 )

Using z table

= 1 - 0.0965

= 0.9035

Probability = 0.9035