STAT500 HW#5_solution
Stat 500 Solution- HW5
1) (25 pts; each 5 pts)
a) z = (y –)/σ = (600 – 550)/100 = 0.5. From Table 1, P (y > 600) = P (z >0.5) = 1 – P (z 
0.5) = 1 – 0.6915 = 0.3085. Therefore, 30.85% of scores can be expected to be greater than
600.
b) z = (y – )/σ = (750 – 550)/100 =2. From Table 1, P (y > 750) = P (z > 2) = 1 – P (z  2) =
1 -0.9772 = 0.0228. So 2.28% would be expected to be greater than 750.
c) z = (y –)/σ = (400 – 550)/100 = - 1.5. P (y < 400) = P (z < -1.5) = 0.0668. Thus 6.68% can
be expected to be less than 450.
d) From part a, P (y < 600) = 0.6915; from part c, P (y < 400) = 0.0668. Then P (400 < y <
600) = P (y < 600) – P (y < 400) = 0.6915 - 0.0668= 0.6247. So 62.47% of scores can be
expected to be between 400 and 600.
e)
y 
n
 100
16
 25 ,
P(ȳ < 440) = P(z < {(440-550) / (100/√16)}) = P(z < -4.4) = 0.0000
P(ȳ <580) = P(z < {(580-550) / (100/√16)}) = P(z < 1.2) = 0.8849
P(440 <ȳ < 580) = P(ȳ <580) -P(ȳ < 440) = 0.8849
Therefore, 88.49% of the class average SAT score is between 440 and 580.
2) (0 pts)
Randomly generate 25 integers from 1 to 1725 with replacement from Minitab:
1539 283 61 1639 302 1081 544 396 1124 1153 1171 753 75 450 1334
667 365 476 696 1036 1391 485 1038 1464 552
3) (15 pts)
Individual baggage weight has
P(Y  100)  P( z 
(Alternative method:
Total weight has mean
  97   y
and
 = 30, so  y  30
200
 2.1213 
100  97
)  P( z  1.4142)  1  0.921348  0.078652 or
2.1213
 y  n  
= 200  97 = 19,400 with the standard deviation
 y =
n   = 200  30 = 424.264. So P (  Y > 20,000) = P {z > (20,000 – 19,400)/ 424.264} =
P (z > 1.4142) = 1 – 0.921348 = 0.078652. The probability that the total weight of the passengers’
baggage will exceed the 20000-pound limit is 7.8652%.)
4) (25 pts) (c) extra 3 credits)
a) μ = 155 and σ = 20  P (y<150) = P {z < (150 -155)/20} = P (z < -0.25) = 0.4013. If a
single measurement is taken, the probability that this measurement will fail detect that the
patient has high blood pressure is 40.13%.
b)
 y  155, y  20 / 6  8.165  P ( y < 150) = P {z < (150 – 155)/8.165} =
= P (z < -0.61237) = 0.270146. If six measurements (n=6) are taken, there is 27.01 %
probability that the average of 6 blood pressure readings will be less than 150.
c) From Table 1, closest value to 0.05 is 0.0505/0.0495 with corresponding z value = -1.645
z = -1.645 = (150-155)/(20/ n )  n = 43.2964 44. So at least 44 measurements are
needed.
5) (25 pts)
1
STAT500 HW#5_solution
a) (y-µ) / () = (2240-2280) / 10.6 = -3.773584906
P(y>2240) = P (z> -3.77358) = 1- P(z<-3.77358)=1-0.0000805=0.9999195
b) µ of y = 2280, the standard deviation (σ) of y = 10.6/ 25 = 2.12.
The central limit theorem will hold here since the observations are normally
distributed. This means, the sampling distribution for y based on random samples of
25 1-foot sections is approximately a normal distribution with the parameters above.
6) ( 0 pts)
a) population mean = __16__, population st. dev. = __5
Sample size
mean of sample
mean
st. dev. Of sample
mean
Skew
N=2
16.01
3.54
-0.02
N=5
16
2.25
-0.01
N=25
16.00
0.99
0.02
b) population mean = 8.08 , population st. dev. = _6.22
Sample size
mean of sample
mean
st. dev. of sample
mean
skew
N=2
8.06
4.36
0.56
N=5
8.13
2.79
0.32
N=25
8.08
1.28
0.20
c) The central limit theorem that for a large sample size (rule of thumb: n  30), y
is approximately normally distributed, regardless of the distribution of the population
one samples from. If the population has mean  and standard deviation  , then y
has mean  and standard deviation 
n
.
The simulation results indicate the central limit theorem seem correct. In (a), the population
distribution is a normal. The sample mean is close to the population mean for different sample size
N . The standard deviation decreases in proportion of1 / N . The skew is close to 0. Hence, the
sample distribution seems to be close to a normal distribution. In (b), the population distribution is
a skewed one. The sample is getting close to the population mean. The standard deviation also
decreases in proportion of 1 /
N . The skew is high (0.56) for small N , but it is getting close to 0
for large N . In this case, its sample mean seems approximately normally distributed. Overall, the
central limit theory holds true regardless of the population distribution as long as the sample size
gets large.
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