If X is normally distributed and P(X<20)=0.5 and P(X>26)=0.0228, find the mean and standard deviation of X. (Make it readable)
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A: mean = 100 s = 15
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If X is
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- Let, X has a normal distribution with a mean of 60 and a standard deviation of 4. The z-value for x = 53 is?A distribution of values is normal with a mean of 120 and a standard deviation of 6. Find the interval containing the mid-44% of the values. Please show your answer to 1 decimal place. Interval = ( , )Biological oxygen demand (BOD) is a pollution index that is monitored in the treated effluent of paper mills. A certain paper mill has taken 31 samples over the past 3 months. The mean and standard deviation for the sample data are 3.42 and 0.728 ppm respectively. The mill would like to see if this sample indicates that the true average BOD in their treated effluent exceeds the targeted 3.00 ppm. Let ? = 0.05. A) The test statistic, (ttest), for this data set is: Note: Round your answer to the nearest hundredth.B) The p-value is: C) The statistical decision and corresponding English interpretation for this study are: Fail to reject Ha: conclude that the true BOD in treated effluent exceeds 3 ppmFail to reject Ho: we cannot conclude that the true BOD in treated effluent exceeds 3 ppm Reject Ha: we cannot conclude that the true BOD in treated effluent exceeds 3 ppmReject Ho: conclude that the true BOD in treated effluent exceeds 3 ppm
- Shoulder height is known to be normally distributed with a mean of 56.65 inches for males and 52.5 inches for females. The males vary with a standard deviation of 2.46 inches and females at 2.49 inches. find the z-score representing a male whose shoulder height is 49 inches a standardized curve N(0,1) is shown below use this information to find an equivalent shoulder height for females and explain the relevance of this value and shading in the context of the problemAssume x is normally distributed with a mean of 2 and a standard deviation of 2. Find: a.P(X=2) b.P(X<2) c.P(X>=2) d.P(2 < X < 4)A distribution of values is normal with a mean of 60 and a standard deviation of 3. Find the interval containing the mid-68% of the values. Please show your answer to 1 decimal place. Interval = (,)
- IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual. (a) Find the z-score for an IQ of 99, rounded to three decimal places. (b) Find the probability that the person has an IQ greater than 99. (c) Shade the area corresponding to this probability in the graph below. (Hint: The x- axis is the z-score. Use your z-score from part (a), rounded to one decimal place). Shade: Left of a value -1 -1.5 0 +). Click and drag the arrows to adjust the values. 1 2 (d) MENSA is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the MENSA organization. (e) Sketch the graph, and write the probability statement. Edit Insert Formats B I U x₂ x² A => EM e & N 18 (f) The middle 50% of IQs fall between what two values? (g) Sketch the graph and write the probability statement. Edit Insert Formats BIUX₂ X² A E = = E EC P R N • Σ+ Σ Α Σ+ Σ ΑSuppose the mean cholesterol levels of women age 45-59 is 5.2 mmol/l and the standard deviation is 0.7 mmol/l. Assume that cholesterol levels are normally distributed. Find the probability that a woman age 45-59 has a cholesterol level above 6.1 mmol/l (considered a high level). Round to four decimal places.P(x > 6.1) = Suppose doctors decide to test the woman’s cholesterol level again and average the two values. Find the probability that this woman’s mean cholesterol level for the two tests is above 6.1 mmol/l. Round to four decimal places.P(x̄ > 6.1) = Suppose doctors being very conservative decide to test the woman’s cholesterol level a third time and average the three values. Find the probability that this woman’s mean cholesterol level for the three tests is above 6.1 mmol/l. Round to four decimal places.P(x̄ > 6.1) =If X is normally distributed and P(X<20)=0.5 and P(X>26)=0.0228, find the mean and standard deviation of X. Give a detailed solution
- Muhammad Akram wishes to borrow Rs 90,50,000 from his bank in order to construct a home. The loan officer does not have any particular statistics on the bank's history of home loans, but he does inform Muhammad Akram that throughout the years, the bank has received around 900 loan applications every year, with an average approval probability of 0.85. Question: Assume that after extensive study, the loan officer informs Muhammad Akram that the correct statistics are 1,076 applications each year with a likelihood of approval of 0.77. What is the current mean and standard deviation?Now use the binomial formulas below to find the mean and standard deviation when n = 5 and p = 0.04. Mean = u = n x p Standard Deviation = q = _/npqSuppose the average life of kiwi phones is 4.66 years with standard deviation 1.05 years. Let X represent the average lif of 40 r/s kiwi phones. Find P( x > 5)
