Question

Suppose a random sample of n = 5 observations is selected from a population that is normally distributed, with mean equal to 3 and standard deviation equal to 0.35.

(a)

Give the mean and standard deviation of the sampling distribution of x. (Round your standard deviation to four decimal places.)

(b)

Find the probability that x exceeds 3.3. (Round your answer to four decimal places.)

(c)

Find the probability that the sample mean x is less than 2.5. (Round your answer to four decimal places.)

(d)

Find the probability that the sample mean deviates from the population mean by more than 0.4. (Round your answer to four decimal places.)

μ = 3

Answer 1

Solution :

Given that,

mean = = 3

standard deviation = = 0.35

n = 5

( a )

= 3

= / n = 0.35 / 5 = 0.1565

The sampling distribution of the sample mean = 3

The standard deviation of the sample mean = 0.1565

( b )

P( > 3.3 ) = 1 - P( < 3.3 )

= 1 - P[( - ) / < ( 3.3 - 3 ) / 0.1565 ]

= 1 - P(z < 1.92)

Using z table,    

= 1 - 0.9726

= 0.0274

probability = 0.0274

( c )

P( < 2.5 )

= P[ ( - ) /   < ( 2.5 - 3) / 0.1565 ]

= P(z < -3.19)

Using z table,

= 0.0007

probability = 0.0274

( d )

The z - distribution of the 0.4 is,

P(Z > z) = 0.4

= 1 - P(Z < z ) = 0.4

= P(Z < z ) = 1 - 0.4

= P(Z < z ) = 0.6

= P(Z < 0.253) = 0.6

z = 0.253

Using z-score formula  

= z * +

= 0.253 * 0.1565 +3

= 3.0396

Answer = 3.0396