Question

The mean balance that college students owe on their credit card is $1996 with a standard deviation of $350. If all possible random samples of size 169 are taken from this population, determine the following:

a) name of sampling distribution

b) mean and standard error of sampling distribution of the mean (use the correct name and symbol for each)

c) percent of sample means for a sample of 169 college students that is greater than $2000

d) probability that sample means for samples of size 169 fall between $1950 and $2050

e) Below which sample mean can we expect to find the lowest 25% of all the sample means?

Answer 1

Solution :

Given that,

mean = = 1996

standard deviation = = 350

n = 169

a) The sampling distribution is normal distribution.

= 1996

(b) Standard error = / n = 350 / 169 = 26.9231

The sampling distribution of the mean = 1996

The standard error of sampling distribution of the mean = 26.9231

(c)

P( > 2000 ) = 1 - P( < 2000)

= 1 - P[( - ) / < ( 2000 - 1996) / 26.9231]

= 1 - P(z < 0.15)

Using z table,    

= 1 - 0.5596

= 0.4404

= 44.04%

Answer : = 44.04%

d)

P( 1950 < < 2050 )  

= P[( 1950 - 1996) / 26.9231 < ( - ) / < ( 2050 - 1996) / 26.9231 )]

= P( -1.71 < Z < 2.01 )

= P(Z < 2.01) - P(Z < -1.71 )

Using z table,  

= 0.9778 - 0.0436  

= 0.9342

Probability = 0.9342

e)

The z distribution of the 25% is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.6745

Using z-score formula  

= z * +

= -0.6745 * 26.9231 + 1996

= 1977.84

Answer : = 1977.84