Download STAT 161 Fall 15

STAT 161 Fall 15
Assignment 5 Solutions (Total marks: 75)
1.134 (4 marks – 2 for z-scores, 2 for comment)
Jessica has a z-score of (1825 – 1498)/316 = 1.03. Ashley’s z-score is (28 – 21.5)/5.4 = 1.20. Ashley’s score
is higher.
1.144
(a) (3 marks) z = (40 − 55)/15.5 = −0.97. From Table A, 16.6% of women have low values of HDL.
(b) (3 marks) z = (60 – 55)/15.5 = 0.32. From Table A, 37.45% of women have protective levels of HDL.
(c) (4 marks) (1 – 0.3745) – 0.1660 = 0.4595. 45.95% of women are in the intermediate range for HDL.
1.145
(a) (3 marks) z = (40 − 46)/13.6 = −0.44. From Table A, 33% of men have low values of HDL.
(b) (3 marks) z = (60 – 46)/13.6 = 1.03. From Table A, 15.15% of men have protective levels of HDL.
(c) (4 marks) (1 – 0.1515) – 0.33 = 0.5185. 51.85% of men are in the intermediate range for HDL.
1.175 (6 marks – explanations are fine, no examples required) No, and no: It is easy to imagine
examples of many different data sets with mean 0 and
standard deviation 1, for example, {−1,0,1} and {−2,0,0,0,0,0,0,0,2}. Likewise, for any given five numbers
a ≤ b ≤ c ≤ d ≤ e (not all the same), we can create many data sets with that five-number summary, simply by
taking those five numbers and adding some additional numbers in between them, for example (in increasing
order): 10, , 20, , , 30, , , 40, , 50. As long as the number in the first blank is between 10 and 20, and so on,
the five-number summary will be 10, 20, 30, 40, 50.
5.4 (6 marks) With n = 49, the distribution of x is approximately N(185, 10). The mean is the same as
the population mean, and the standard deviation is
σ
70 70
σ x= =
= =10
√ n √ 49 7
About 95% of the time, ̄x is between 165 and 205 (two standard deviations either side of the mean).
5.10 (a) (2 marks) The population mean, is considered fixed. It does not have a distribution. The
distribution of sample means will be approximately Normal with large n.
(b) (2 marks) The distribution of the observed values will “look like” the population distribution, and it is
the distribution of the sample mean that becomes approximately Normal for large n.
2σ
(c) (3 marks) For large n, ̄x will be within μ±2σ x = μ±
about 95% of the time, not within
√n
μ±2σ / n .
5.14 (a) (2 marks)
σ
1.24
σ x= =
=0.101
√ n √ 150
(b) (3 marks) For 95% of samples, we’ll expect the sample mean to be within
between 6.578 and 6.982 hours.
6.78±2(0.101) hours, or
(c) (4 marks) Using Table A, the probability is 0.8830.
σ 0.82
=
=0.205 .
√ n √ 16
5.18 (a) (2 marks) n = 40 is generally considered “large enough” for the sample mean to be approximately
Normal.
(b) (4 marks) The graph of the approximate distribution is given below. The standard deviation is = 0.130.
5.16 (a) (2 marks) The standard deviation will be
σ x=
(c) (4 marks)
̄ <7.25∨ X
̄ >7.25) By symmetry, this is
We need P ( X
̄ <7.25)=2 P ( Z < 7.25 )=2 P( Z <−1.15) .Using Table A, the desired probability is 2(0.1251)
2 P(X
0.13
=0.2502.
(11 marks: 3 for (a), 3 for (b), 3 for (c), 2 for (d))
μσ