Download Exam 2 - RIT

SMAM 314 Exam 2 Name_______________ 1. The strength of an aluminum alloy is normally distributed with mean 10 gigapascals (GPa) and standard deviation 1.6 GPa. A. What is the probability that a specimen of this alloy will have a strength stronger than 12 GPa? (8 points) ⎛
12 − 10 ⎞
P(X > 12) = P ⎜ Z >
= P(Z > 1.25) = .1056 1.6 ⎟⎠
⎝
normalcdf(12,10^99,10,1.6) = .1056 B. What will be the strength of the weakest specimen of this alloy among the strongest 5% of the alloys?(8 points) P(X > c) = .05
⎛
c − 10 ⎞
P⎜ Z >
= .05
1.6 ⎟⎠
⎝
c − 10
= 1.645
1.6
c = 10 + 1.645(1.6) = 12.632
invNorm(.95,10,1.6)=12.632 2. Use the Central Limit Theorem to do this problem. The breaking strength in kg/mm for a certain type of fabric has mean 1.86 and standard deviation 0.27. A random sample of 80 pieces of fabric is drawn. What is the probability that the sample mean breaking strength is less than 1.8 kg/mm? (12 points) ⎛
1.8 − 1.86 ⎞
P(x < 1.8) = P ⎜ Z <
⎟ = P(Z < −1.99) = .0233 ⎝
.27 / 80 ⎠
nomalcdf(‐10^99, 1.8,1.86,.27/sqrt[80])=.0234 3. Use the normal approximation to the binomial distribution with the continuity correction. Among the adults in a large city 30% have a college degree. A simple random sample of 100 adults is chosen. What is the probability that between 25 and 35 adults inclusive have a college degree?(12 points) µ = 100(.30) = 30
σ = 100(.30)(.70) = 4.582
⎛ 24.5 − 30
35.5 − 30 ⎞
P(25 ≤ X ≤ 35) = P ⎜
<Z<
= P(−1.20 < Z < 1.20) = .7699
⎟
4.582 ⎠
⎝ 4.582
normalcdf (24.5,35.5,30,4.582) = .7700
4. A random variable X has the continuous probability density ⎧⎪ 6x(1 − x) 0 < x < 1
f (x) = ⎨
0
elsewhere
⎩⎪
Find A. P(.25<X<.75) (5 points) P(.25 < X < .75) =
.25
6x(1 − x)dx = .6875 B. The mean of X(5 points) EX =
∫
.75
1
∫ 6x
0
2
(1 − x)dx = .5 C. The standard deviation of X.(5 points) EX 2 =
1
∫ 6x (1 − x)dx = .3
3
0
σ 2 = .3 − (.5)2 = .05
σ = .2236
5. The time T between requests to a web server in seconds is exponentially distributed. The cumulative distribution function is F(t) = 1 − e −2t , t > 0 A. For what value of t is P(T ≤ t) = 0.5? (6 points) 1 − e −2t = .5
e −2t = .5
−2t = ln(.5)
t = −.5ln(.5) = .3466
B. Without doing any integration what is the expected time between requests to the server?(3 points) .5 sec C. Suppose that for the last two seconds there have not been any requests. What is the probability that more than one additional second will elapse before the next request?(6 points) 1 − F(1) = 1 − (1 − e −2(1) ) = e −2 = .135 6. The capacities in ampere‐hours were measured for a sample of 120 batteries. The average was 178 and the standard deviation was 14. A. Find a 95% confidence interval for the mean capacity of batteries produced by this method.(6 points) zσ
x±
n
1.96(14)
178 ±
120
178 ± 2.505
(175.495,180.505)
B. Interpret the confidence interval obtained in part A.(3 points) The probability that the true mean battery capacity lies in the interval is .95. C.Approximately how many batteries must be sampled so that a 95% confidence interval will specify the mean to within ±2 ampere hours? 1.96(14)
2=
n
n = 1.96(7)
n = (1.96(7))2 = 188.23
n = 189
7.A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average‐system activation temperature is 130 degrees F. A sample of nine systems ,when tested, yields a sample average activation temperature of 131.08 degrees. It is known that the distribution of activation temperatures is normal with standard deviation 1.5 degrees F. A test of hypothesis is performed in order to determine whether the average system activation temperature is significantly different from 130 degrees F. For this test of hypothesis (14 points‐2 per part) A. What is the null and the alternative hypothesis? H0 µ=130
H1 µ≠130
B. What assumptions are made to perform this test? Normal population, known standard deviation C. What is the region of rejection at α=.05?
Ζ>1.96 Ζ<−1.96
D. What is the value of the test statistic? Z=
Z=
x−µ
σ/ n
131.08 − 130
1.5 / 9
= 2.16
E.At α=.05 would you reject or fail to reject H0? Reject H0 at α=.05 . F. What would you say to your boss about the average system activation temperature? There is strong evidence that the average system activation temperature is different from 130. G.What is the p value? P(Z>2.16)+P(Z<–2.16)=.0308 H. Based on this p value would you reject or fail to reject H0 at α=.01? Explain. Fail to reject H0 because the p value is greater than .01.