Homework help

)In a certain year, when she was a high school senior, Idonna scored 670 on the mathematics part of the SAT. The distribution of SAT math scores in that year was Normal with mean 515 and standard deviation 113. Jonathan took the ACT and scored 21 on the mathematics portion. ACT math scores for that year were Normally distributed with mean 20.9 and standard deviation 4.7. Find the standardized scores for both students. (Round your answers to two decimal places.)Idonna =z = (x-µ)/sdz =(670-515)/113   = 1.37Jonathan = z = (x-µ)/sd(21-20.9)/4.7 = 0.022)Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). The exam is composed of three multiple-choice sections (Physical Sciences, Verbal Reasoning, and Biological Sciences). The score on each section is converted to a 15-point scale so that the total score has a maximum value of 45. The total scores follow a Normal distribution, and in a certain year the mean was 25.2 with a standard deviation of 6.9. There is little change in the distribution of scores from year to year. (Round your answers to four decimal places.)(a) What proportion of students taking the MCAT had a score over 32?  For the proportion of student who scored  above 32 isX> 32  which corresponds with  Z > (32- 25.2)/6.9 = 0.99from a standard normal distribution the probability of a z value of less or equal to 0.99 is 0.8389so  P(X>32) =1-0.8389= 0.1611so the PROPORTION which scored above 32 is 0.1611(b) What proportion had a score between 22 and 25?  Z (22) =  (22- 25.2)/6.9=  -0.46Z(25)  = (25- 25.2)/6.9 = -0.02we are checking the probability of lying between-0.02> Z > -0.46=.4920 – 3228 =0.16923)The total scores on the Medical College Admission Test (MCAT) follow a Normal distribution with mean 24.8 and standard deviation 6.9.(a) What are the median and the first and third quartiles of the MCAT scores? (Enter your median to one decimal place. Round your quartiles to the nearest whole number.)median=The  median  is 50% point of the data which is a probability 0.5 From standard normal distribution it corresponds to  a z of 0.000.0     =( x –24.8)/6.9 x = 24.8Again a normal distribution is symmetrical with mean at the centerQ1= this is the 0.25 of the left tailThe z value for this  probability is  -0.67So z= (x- 24.8)/6.9 =-0.67-4.623 = x-24.8X = -4.23+24.8 =20Q3= this is the 0.75  of the data The z value for this  probability is 0.67Z =(x- 24.8)/6.9 =0.674.623 +24.8=29What is the interquartile range? (Round your answer to the nearest whole number.)  The interquartile  range is  upper quarter – lower quater                     29-20 =9(b) Give the interval that contains the central 80% of the MCAT scores. (Round your answers to the nearest whole number.)  This can be obtained by taking 90th perventile – 10th percentileFor the 90th we find z  corresponding to probability 0.9 This is 1.28 from an std normal table  1.28=( x -24.8)/6.98.832 =x-24   X= 32.832  =33For 0.10  corresponds to -1.29-1.29 =( x-24.8)/6.9 -8.901 =x-24.8X= 15.899 = 16The interval is 16 -324)It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minutes per day that people spend standing or walking. Among mildly obese people, minutes of activity varied according to the N(378, 68) distribution. Minutes of activity for lean people had the N(526, 106) distribution. Within what limits do the active minutes for 95% of the people in each group fall? Use the 68-95-99.7 rule. Within what limits do the active minutes for 95% of the people in the mildly obese group fall?Mean =378 sd =68It  lies between 2.5th quartile and 97.5th quartile The z value for  0.975 is 1.96  1.96= (x-378)/68133.28= x-378X= 133.28+ 378 = 511.28For 2.5th quartileZ =-1.96-133.28 =x-378X= 244.72(  244.72    ) to (    511  ) minWithin what limits do the active minutes for 95% of the people in the lean group fall?  It  lies between 2.5th quartile and 97.5th quartileMean =526,sd =106 The z value for  0.975 is 1.96  1.96= (x-526)/106207.76= x-526X= 207.76+ 526 = 733.76For 2.5th quartileZ =-1.96-1.96= (x-526)/106-207.76= x-526X= -207.76+ 526 = 318.24(    318 ) to (   734  ) min5)The common fruit fly Drosophila melanogaster is the most studied organism in genetic research because it is small, is easy to grow, and reproduces rapidly. The length of the thorax (where the wings and legs attach) in a population of male fruit flies is approximately Normal with mean 0.810 millimeters (mm) and standard deviation 0.083 mm. (Round your answers to four decimal places.)(a) What proportion of flies have thorax lengths less than 0.71 mm? Mean =0.810 ,sd=0.083The z corresponding to length 0.71  = 0.55Z =(x-u)/sd=(0.71-0.810)/0.083=-1.2048193From the table it is0.1151(b) What proportion have thorax lengths greater than 0.91 mm? The z corresponding to probability 0.09  is  -1.34 Z= ( 0.91-0.810)/0.083=1.2048193From the table=1-0.8849 ==0.1151(c) What proportion have thorax lengths between 0.71 mm and 0.91 mm? Proportion corresponding to 0.91 = P(0.91) –P(0.71)= 0.8849-0.1151=   0.76986)In a study of exercise, a large group of male runners walk on a treadmill for 6 minutes. Their heart rates in beats per minute at the end vary from runner to runner according to the N(106, 12.6) distribution. The heart rates for male nonrunners after the same exercise have the N(139, 17) distribution. (Round your answers to two decimal places.)(a) What percent of the runners have heart rates above 143?  Z =(143-106)/12.6=2.9365079From an std normal  table we have0.9984Prop Above 143 is 1-0.9984 =0.0016(b) What percent of the nonrunners have heart rates above 143? Z =(143-139)/17=0.24From an std normal  table we have0.5948Prop Above 143 is 1-0.5948 =0.4052