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6.6.q7 - Student Grady Simonton Course Math119 Elementary...

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Unformatted text preview: Student: Grady Simonton Course: Math119: Elementary Statistics - Spring 2.010 - CRN: 49239 Instructor: Shawn Parvini - 16 weeks Date: 3/18/10 Book: Triola: Elementary Statistics, 11e Time: 4:16 PM (a) With n = 13 and p = 0.4, find P(at least 9) using a binomial probability table. (b) If np 2 5 and nq 2 5, also estimate P(at least 9) by using the normal distribution as an approximation to the binomial distribution; if np < 5 or nq < 5, then state that the normal approximation is not suitable. (a) Where n = 13 and p = 0.4, find P(at least 9). The binomial probability of at least 9 is the sum of the probabilities for values of x ranging from 9 to n. P(at least 9) P(9) +P(10) +P(ll) +P(12) +P(l3) .024 +006 +00] +0 +0 0.03] (b) The normal distribution is suitable for approximating the binomial probability when both np and nq are greater than or equal to 5. Determine whether the normal distribution is suitable for approximating this binomial probability. Start by finding the value of q, the probability that the event will not occur. Recall p = 0.4. q=1-p =1—0.4 =0.6 Find the values ofnp and nq. Recall n = 13, p = 0.4, and q = 0.6. 13 ' 0.4 ml 5.2 13 ' 0.6 7.8 “P Because np = 5.2 and nq = 18 are both greater than or equal to 5, the normal distribution is suitable for approximating the binomial probability. Find the values of u and U for the normal distribution. Recall n = 13, and p = 0.4, q = 0.6. u=n-p U=1in'p'q (13)(0.4) = V 13 - 0.4 - 0.6 5.2 = 1.766 The probability of interest, P( at least 9), is shown at the right. Because normal probabilities are found for areas to the left of a specific value, calculate P(at least 9) as 1 - P(x < 9). x Using a continuous distribution to approximate a discrete probability, x < 9 needs to be adjusted for continuity. Because the inequality is less than, adjust x by subtracting 0.5. Page 1 Student: Grady Simonton Course: Math119: Elementary Statistics - Spring 2.010 - CRN: 492.39 Instructor: Shawn Parvini - 16 weeks Date: 3/18/10 Book: Triola: Elementary Statistics, 11e Time: 4:16 PM x < 9 — 0.5 = 8.5 Find the z score corresponding to x = 8.5. Recall that u = 5.2 and o = 1.766. (9 — 0.5)—5.2 295 —= l. 7 1.766 8 Approximate P(x S 8), the area of the 3’ normal curve to the left of 8.5. Pass) = P(z< 1.87) = 0.9693 g_5 Determine P(at least 9), the area of the normal curve to the right Y of 8.5. P(at least 9) = l - P(x < 9) x = 1 — 0.9693 8.5 = 0.0307 Appropriate table values were used to determine both the binomial probability and the probability using the normal distribution as an approximation to the binomial. Notice that the difference between the binomial probability, 0.03 l , and the approximation using the normal distribution, 0.0307, is small. Page 2. ...
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