The mean balance that college students owe on their credit card is $1096 with a

Question

The mean balance that college students owe on their credit card is $1096 with a standard deviation of $150. If all possible random samples of size 169 are taken from this population, determine the following: a) name of the Sampling Distribution b) mean and standard error of the 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 are greater than $1120. d) probability that sample means for samples of size 169 fall between $1080 and $1110. e) Below which sample mean can we expect to find the lowest 25% of all the sample means?

Explanation / Answer

a)
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
b)
mean of the sampling distribution ( x ) = 1096
standard Deviation ( sd )= 150/ Sqrt ( 169 ) =11.5385
sample size (n) = 169
c)
P(X > 1120) = (1120-1096)/150/ Sqrt ( 169 )
= 24/11.538= 2.08
= P ( Z >2.08) From Standard Normal Table
= 0.0188
d)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 1080) = (1080-1096)/150/ Sqrt ( 169 )
= -16/11.53846
= -1.38667
= P ( Z <-1.38667) From Standard Normal Table
= 0.08277
P(X < 1110) = (1110-1096)/150/ Sqrt ( 169 )
= 14/11.53846 = 1.21333
= P ( Z <1.21333) From Standard Normal Table
= 0.8875
P(1080 < X < 1110) = 0.8875-0.08277 = 0.80473

e)
P ( Z < x ) = 0.25
Value of z to the cumulative probability of 0.25 from normal table is -0.67449
P( x-u/s.d < x - 1096/11.5385 ) = 0.25
That is, ( x - 1096/11.5385 ) = -0.67449
--> x = -0.67449 * 11.5385 + 1096 = 1088.2174

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