Question

Consider the following sample of observations on coating thickness for low-viscosity paint.

0.83		0.88		0.88		1.07		1.09		1.15		1.29		1.31
1.34		1.49		1.59		1.62		1.65		1.71		1.76		1.83

Assume that the distribution of coating thickness is normal

(a) Calculate a point estimate of the mean value of coating thickness.

1.343125

(b) Calculate a point estimate of the median of the coating thickness distribution.

1.325

(c) Calculate a point estimate of the value that separates the largest 10% of all values in the thickness distribution from the remaining 90%.

1.761634 WRONG

(d) Estimate P(X < 1.5), i.e., the proportion of all thickness values less than 1.5. [Hint: If you knew the values of ? and ?, you could calculate this probability. These values are not available, but they can be estimated.] (Round your answer to four decimal places.)
0.6949 WRONG

(e) What is the estimated standard error of the estimator that you used in part (b)? (Round your answer to four decimal places.)
0.085176 WRONG

I've been following the examples here in chegg and for some reason im not able to get them right... please write down the steps clearly.

I've been using excel and I am not sure what is wrong, the values I got are the following:

1) S I got from part c is S=0.340707597, x=1.761634354 which was marked wrong.

2) Z<1.5=0.69497 from part d also marked wrong

3) SE=0.085176899 from part e also marked wrong

Answer 1

a)
mean = 1.343125

b)
In order to find the median, arrange the data in ascending order and take the average of 8th and 9th number.
median = (1.31 + 1.34)/2 = 1.325

c)
mean = 1.343125
std. dev. = 0.3338 (Excel formula to calculate std. dev. : STDEV.S())

For 0.1, z-value = 1.2816 (Excel formula to calculate z-value =NORM.INV(0.9,0,1))

using central limit theorem,
x = 1.343125 + 1.2816*0.3338
x = 1.7709

d)
P(X < 1.5)
= P(z < (1.5 - 1.343125)/(0.3338/sqrt(16)))
= P(z < 1.8799)
= 0.9699

e)
std. error = (0.3338/sqrt(16))
se = 0.08345