If the random variable X is normally distributed with a mean equal to .45 and a standard deviation equal to .40, then P(X ≥ .75) is: (a) 0.2266 (b) 0.4525 (c) 0.7734 (d) 0.7500 (e) None of the above
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- Let x be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the x distribution is ? = 7.4.† A new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 41 patients with arthritis took the drug for 3 months. Blood tests showed that x = 8.5 with sample standard deviation s = 3.0. Use a 5% level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood. (a) What is the level of significance? State the null and alternate hypotheses. (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. What is the value of the sample test statistic? (c) Estimate the P-value. Sketch the sampling distribution and show the area corresponding to the P-value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the…Let x be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the x distribution is ? = 7.4.† A new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 36 patients with arthritis took the drug for 3 months. Blood tests showed that x = 8.6 with sample standard deviation s = 3.0. Use a 5% level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood. (A) What is the value of the sample test statistic? (Round your answer to three decimal places.) (B) Find the P-value. (Round your answer to four decimal places.)Assume the random variable x is normally distributed with mean μ=50 and standard deviation σ=7. Find the indicated probability. P(x> 36)
- The random variable X has the nomal distribution with mean 4.8 andi standard deviation 0.3. Find P(X > 5.4) 0.0228 () 0.2743 0.7257 0.9772 Tpcint.If XX represents a random variable coming from a normal distribution and P(X<10.2)=0.22P(X<10.2)=0.22, then P(X>10.2)=0.78P(X>10.2)=0.78. true or falseLet x be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the x distribution is ? = 7.4.† A new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 31 patients with arthritis took the drug for 3 months. Blood tests showed that x = 8.8 with sample standard deviation s = 3.1. Use a 5% level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood. a) What is the level of significance? b) What is the value of the sample test statistic? (Round your answer to three decimal places.)
- X is a normally distributed random variable with a mean of 5.00 and a standard deviation of 11.39. If the right area is 0.4716 then what is the value of X?A binomial distribution has p = 0.35 and n = 100. If we are trying to approximate its probability using a normal approximation (applying the Central Limit Theorem) what is the standard deviation that we will use for this approximation? (a) 0.4770 (b) 0.2275 (c) 0.0477 (d) 4.7697 (e) None of the above5. Let x be a random variable that represents the level of glucose in the blood (milligrams per deciliter of blood) after a 12-hour fast. Assume that for people under 50 years old, x has a distribution that is approximately normal, with mean u = 85 and estimated standard deviation a = 25. A test result x< 40 is an indication of severe excess insulin, and medication is usually prescribed. (a) What is the probability that, on a single test, x < 40? That's find P(x < 40) (b) Suppose a doctor uses the average x fot two tests taken about a week apart. What is the probabilit that ( Hint: Your standard deviation here should be ) P(ĩ < 40)
- Let x be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the x distribution is ? = 7.4.† A new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 36 patients with arthritis took the drug for 3 months. Blood tests showed that x = 8.6 with sample standard deviation s = 3.5. Use a 5% level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood. What is the level of significance?Fawns between 1 and 5 months old in Mesa Verde National Park have a body weight that is approximately normally distributed 25.36 kg and standard deviation q = 4.72 kg. Let x be the weight of a fawn in kg. What is the probability that for a fawn chosen at random: with mean u - (a) x is less than 28.88 kg? (b) x is greater than 17.34 kg? (c) x lies between 29.04 and 33.6 kg? -