Question

- 07 sampling 's for review-SET 1 FOR POSTING.pdf 0 3 . 12 QUESTION 1 - TESTING WEATHER STATIONS FOR MARS A company has signed a contract to develop a "weather station” for use on Mars. It will continually measure temperature and wind speed. It must work in extreme cold and in the presence of fine dust, for one Martian year. The company wants to test its design, by building several prototypes, and subjecting the prototypes in a lab to conditions similar to Mars. The company claims the probability is 80% that a weather station will perform satisfactorily. The company builds and tests ten prototypes. These prototypes are built and tested independently. If the company's claim is correct that the probability of a weather station performing satisfactorily is 80%, what is the probability at least eight of the prototypes will perform satisfactorily? What sampling distribution should we use to compute this probability? a. binomial b. normal c. approximately normal (CLT) d. t e. chi-square

Answer 1

1)

We will use binomial distribution.

The population proportion of success is p = 0.8, also, 1-p=1-0.8 = 0.2, and the sample size is n = 10. We need to compute Pr(X > 8) Therefore, we get that Po(X 2 9) = Pax = 5) = § () »(1 –p)"; i=8 This implies that Pr(X > 8) = Pr(X = 8) + Pr(X = 9) + Pr(X = 10) = (5) 0.8 x 0.310-8 + (19) 0.8" x 0.20-9 + (10) 0.310 x 0.210 10 = 0.302 +0.2684 + 0.1074 = 0.6778 which means that the probability we are looking for is Pr(X > 8) = 0.6778.

2) We will use the binomial distribution

The following information is provided: The population proportion of success is p=0.9, also, 1-p=1 – 0.9 = 0.1, and the sample size is n = 100. We need to compute Pr(85 < X < 95) Therefore, we get that P1(86 < X < 0) = P(X=1 = F (*) (1 - mit i=85

This implies that Pr(85 < X <95) = Pr(X = 85) + Pr(X = 86) + Pr( X = 87) + Pr(X = 88) + Pr(X = 89) + Pr( X = 90) + Pr(X = 91) + Pr(X =92) + Pr( X = 93) + Pr(X = 94) + Pr(X = 95) 100 0.985 x 0.1100-85+ (100) 0.986 x 0.1100–86 + 86 100 87 0.987 x 0.1100-87 + 0.988 x 0.1100-88 + 100 89/ 0.989 x 0.1100–89 + 100 90 0.990 x 0.1100-90+ (1000.991 x 0.1100–91 + 91 = 0.0327 + 0.0513 +0.0743 + 0.0988 + 0.1199 +0.1319 +0.1304 + 0.1148 +0.0889 + 0.0596 + 0.0339 = 0.9364 which means that the probability we are looking for is Pr(85 < X < 95) = 0.9364.

3)

We will use the t distribution as we have sample statistics, not population statistics.

4)

If the true standard deviation is given that is population standard deviation, we will use the normal distribution to compute this probability.