Question

Please answer number 4 and if possible, upload a picture showing all the necessary steps

The serum cholesterol levels of a population of 12-14 year olds follow a normal distribution with a mean of 162 mg/dl and a standard deviation of 28 mg/dl. Consider randomly selecting 9 individuals of this population, measuring each individual's cholesterol level, and then computing the sample mean of these 9 subjects 3. 1 pt If X is the sample mean cholesterol level of the 9 individuals, what is the mean of the sampling distribution of X? 4. (1 pt What is the standard deviation of the sampling distribution of X? 5. 12 pts What is the shape of the sampling distribution of X? What condition was necessary for the sampling distribution to have this shape? 6. 3 pts Compute the probability that the sample mean of the random sample of n 9 is between 152mg/dl and 172 mg/d, P(152<X< 172).

Answer 1

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^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)

Q3.
mean of the sampling distribution ( x ) = 162

Q4.
sample size (n) = 9
standard Deviation ( sd )= 28/ Sqrt ( 9 ) =9.3333

Q5.
the distribution is given that normal, since the curve is normal and we sampled the size is very
minimal and this does n't effects the distribution is to change the shape

Q6.
BETWEEN THEM
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 152) = (152-162)/28/ Sqrt ( 9 )
= -10/9.33333
= -1.07143
= P ( Z <-1.07143) From Standard Normal Table
= 0.14199
P(X < 172) = (172-162)/28/ Sqrt ( 9 )
= 10/9.33333 = 1.07143
= P ( Z <1.07143) From Standard Normal Table
= 0.85801
P(152 < X < 172) = 0.85801-0.14199 = 0.71602