To obtain: The probability that the claim will be accepted. Given info: The sample size is n=100 and the drug manufacturer claims that a drug cures a rare skin disease is 65% of the time.

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To obtain: The probability that the claim will be accepted.  Given info:  The sample size is n=100 and the drug manufacturer claims that a drug cures a rare skin disease is 65% of the time.

Asma Vang · Accepted Answer

Calculation: The drug cures a rare skin disease p is 0.65. The drug not cures a rare skin disease q is 0.35. The requirements to check whether the normal distribution can be used for approximate the distribution of x, the binomial random variable are np≥5 and nq≥5. That is, Condition 1: np≥5 Condition 2: nq≥5 Condition 1: np≥5 Substitute 100 for n and 0.65 for p in the np. np=100×0.65 =65 &amp;gt;5 Thus, the given requirement np(=65)≥5 is satisfied. Condition 2: nq≥5 Substitute 100 for n and 0.35 for q in the nq. nq=100×0.35 =35 &amp;gt;5 Thus, the given requirement nq(=35)≥5 is satisfied. Since the requirements that np≥5 and nq≥5 both are satisfied. Thus, the normal distribution can be used to approximate the binomial distribution. That is, the probability from a binomial probability distribution can be approximated by using normal distribution with the parameters are, μ−np and σ=npq. The value of mean and standard deviation is, μ=100×0.65 =65 σ=100×0.65×0.35 =22.75 =4.7697 Here, the claim will be accepted represent that the binomial random variable x is at least 70. Continuity correction: If the binomial probability represents &quot;at least c&quot; then the normal probability is P(x&amp;gt;c−0.5) for any number c. By using continuity correction, the value 0.5 is subtracted from the value of 70. That is, c=70 P(x≥70)=P(x&amp;gt;70−0.5) =P(x&amp;gt;69.5) Here, P(x&amp;gt;69.5) represents the area to the right of 69.5. The formula to convert the x value into z score is, z=x−μσ Substitute 69.5 for x, 65 for μ and 4.7697 for σ z=69.5−654.7697 =4.54.7697 =0.94 The probability that the claim will be accepted is obtained by finding area to the right of 0.94. But Table 4, the Standard normal distribution is used to find the areas to the left. Use Table 4: Standard normal distribution to find the area to the left of 0.94. Procedure: For z at 0.94, - Locate 0.9 in the left column of the Table 4. - Obtain the value in the corresponding row below 0.04. That is, P(z&amp;lt;0.94)=0.8264 The area to the right of 0.94 is, P(z&amp;gt;0.94)=1−P(z&amp;lt;0.94) =1−0.8264 =0.1736 Thus, the probability that the claim will be accepted is 0.1736.

To obtain: The probability that the claim will be accepted. Given

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