Statistics 1060: Practice Solutions Week 7

4.61

We know the standard deviation of a distribution is the square root of the variance, $\sigma^{2}_{X}$.
$\sigma^{2}_{X} = (x_{1} - \mu)^{2}p_{1} + (x_{2} - \mu)^{2}p_{2} + ... + (x_{k} - \mu)^{2}p_{k}$

From the practice problems from Week 6, $\mu = 2.25$, so
$\sigma^{2}_{X} = (0 - 2.25)^{2}(0.10) + (1 - 2.25)^{2}(0.15) + (2 - 2.25)^{2}(0.30) +$

(3 - 2.25)²(0.30) + (4 - 2.25)²(0.15) = 1.3875. So $\sigma_{x} = \sqrt{1.3875} = 1.178$.

4.65

Since the two times are independent, the total variance is $\sigma^{2}_{total} = \sigma^{2}_{pos} + \sigma^{2}_{att} = (2)^{2} + (4)^{2} = 20$. So $\sigma_{total} = \sqrt{20} = 4.472$ seconds.

4.69

a) Now a simple calculation, $\mu_{x} = 550^{o}$ C and $\sigma^{2}_{x} = 32.5$. So $\sigma_{x} = 5.701^{o}$ C.

b) Mean = 0^o C, standard deviation = 5.701^o C.

c) $\mu_{y} = \frac{9}{5}\mu_{x} + 32 = 1022^{o}$ F and $\sigma_{y} = \frac{9}{5}\sigma_{x} = 10.26^{o}$ F.

5.3

a) Yes. It is reasonable to assume that the results for the 50 students are independent, and each has the same chance of passing.

b) No. Since the student recieves instruction after incorrect answers, her probability of success is likely to increase.

c) No. Temperature may affect the outcome of the test.

STATISTICS 1060 HOMEPAGE

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