320 Chapter 7 The Normal Distribution Solution The population proportion is p = 0.25 and the sample size is n = 70. Therefore, ��̂p = p = 0.25 ��̂p = √ 0.25(1 − 0.25) 70 = 0.0518 Check Your Understanding 1. Find ��̂p and ��̂p if n = 20 and p = 0.82. ��̂p = 0.82; ��̂p = 0.08591 2. Find ��̂p and ��̂p if n = 217 and p = 0.455. ��̂p = 0.455; ��̂p = 0.03380 Answers are on page 324. The probability histogram for the sampling distribution of p̂ Figure 7.25 presents the probability histogram for the sampling distribution of ̂p for the proportion of heads in five tosses of a fair coin, for which n = 5 and p = 0.5. The distribution is reasonably well approximated by a normal curve. Figure 7.26 presents the probability histogram for the sampling distribution of ̂p for the proportion of heads in 50 tosses of a fair coin, for which n = 50 and p = 0.5. The distribution is very closely approximated by a normal curve. 0 0.2 0.4 0.6 0.8 1.0 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0 Figure 7.25 The probability histogram for ̂p when n = 5 and p = 0.5. The histogram is reasonably well approximated by a normal curve. 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.12 0.10 0.08 0.06 0.04 0.02 0 Figure 7.26 The probability histogram for ̂p when n = 50 and p = 0.5. The histogram is very closely approximated by a normal curve. When p = 0.5, the sampling distribution of ̂p is somewhat close to normal even for a small sample size like n = 5. When p is close to 0 or close to 1, a larger sample size is needed before the distribution of ̂p is close to normal. A common rule of thumb is that the distribution may be approximated with a normal curve whenever np and n(1 − p) are both at least 10. The Central Limit Theorem for Proportions Let ̂p be the sample proportion for a sample size of n and population proportion p. If np ≥ 10 and n(1 − p) ≥ 10 then the distribution of ̂p is approximately normal, with mean and standard deviation ��̂p = p and ��̂p = √ p(1 − p) n
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