45-733 Probability and Statistics I
(3rd Mini AY 1997-98 Flex-Mode and Flex-Time)

Assignment #6 Answers

**7.19** =

=
2*0.9773 - 1 = 0.9546

**7.20**

=

**7.21** We are given: **m** = 5.00,
**s** = .50, and **n** = 64. We are
asked to compute:
**
_
P(X**_{n} £ 4.90) = P[Z £ (4.90 - 5.00)/.5/8] =
P(Z £ -1.6) = F(-1.6) = .0548

**8.40(a)**

** 1-a** = 0.98,
**a** = 0.02,
**a/2** = 0.01,
**Z**_{0.01 }= 2.33

Confidence limits are :

98 % confidence limits are (0.484,0.588)

**(b)** Yes. There is no evidence that the graduation rate has changed.

**8.41**

** 1-a** = 0.99,
**a** = 0.01,
**a/2** = 0.005,
** Z**_{0.005 } = 2.58

Confidence limits are :

98 % confidence limits are (0.784,0.826)

_{^}
**8.43(a)** We are given **p** = 2/3, **n** = 224, **a** = .10, **Z**_{.05} = 1.645
_{^} _{^} _{^}
**p ± Z**_{a/2}[p(1 - p)/n] = 2/3 ± 1.645{[(2/3)(1/3)]/224}^{1/2} =
.667 ± .052 = (.615, .719)

**(b)** Because the entire interval is greater than 1/2, it is reasonable
to believe that most of the children think that they would like to travel
in space.

**8.45** **1-a** = 0.95,
**a** = 0.05,
**a/2** = 0.025,
**Z**_{0.025 } = 1.96,

**m** = 30, **n** = 30

Use large sample confidence interval

or (15.72, 36.677)

**8.51** Large sample CLT proportions

**1-a** = 0.98,
**a** = 0.02,
**a/2** = 0.01,
**Z**_{0.01 }= 2.33

or (-0.05, 0.18)

**8.74(a)** **1-a** = 0.95,
**a** = 0.05,
**a/2** = 0.025,
**t**_{0.025 } = 2.048 (28 degrees of freedom)

or (-120.55, -55.45)

**(b)**

or (-9.8, 71.8)

**(c) ** For the verbal scores, the confidence limits are below 0
indicating that the two populations differ.

For the math scores, 0 is in the interval, so there is not evidence to
conclude that there is a difference.

**(d) ** Must assume samples are drawn from a NORMAL distribution.

**8.86** From the data stated in the problem we can calculate:

**s**^{2} = 144.5, **a** = .01,
**a/2** = .005,
**C**_{1} = .20699, **C**_{2} = 14.8602.

The 99% confidence interval for **s**^{2} is:

**P[(n - 1)s**^{2}]/C_{2} < s^{2} <
[(n - 1)s^{2}]/C_{1}] = .99

And the endpoints of the interval are: (4*144.5)/14.8602 = 38.9 and

(4*144.5)/.20699 = 2792.41

which produces the interval for the standard deviation of:

6.24 < **s** < 52.84