The SAT scores have an average of 1200 with a standard deviation of 60. A sample of 36 scores is selected. a) What is the probability that the sample mean will be larger than 1224? b) What is the probability that the sample mean will be less than 1230? c) What is the probability that the sample mean will be between 1200 and 1214? d) What is the probability that the sample mean will be greater than 1200? e) What is the probability that the sample mean will be larger than 73.46?

Answer:a) 0.0082b) 0.9987c) 0.9192d) 0.5000e) 1Step-by-step explanation:The question is concerned with the mean of a sample.  From the central limit theorem we have the formula:[tex]z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]a) [tex]z=\frac{1224-1200}{\frac{60}{\sqrt{36} } }=2.40[/tex]The area to the left of z=2.40 is 0.9918The area to the right of z=2.40 is 1-0.9918=0.0082[tex]\therefore P(\bar X\:>\:1224)=0.0082[/tex]b) [tex]z=\frac{1230-1200}{\frac{60}{\sqrt{36} } }=3.00[/tex]The area to the left of z=3.00 is 0.9987[tex]\therefore P(\bar X\:<\:1230)=0.9987[/tex]c) The z-value of 1200 is 0The area to the left of 0 is 0.5 [tex]z=\frac{1214-1200}{\frac{60}{\sqrt{36} } }=1.40[/tex]The area to the left of z=1.40 is 0.9192The probability that the sample mean is between 1200 and 1214 is [tex]P(1200\:<\:\bar X\:<\:1214)=0.9192-0.5000=0.4192[/tex]d) From c) the probability that the sample mean will be greater than 1200 is 1-0.5000=0.5000e) [tex]z=\frac{73.46-1200}{\frac{60}{\sqrt{36} } }=-112.65[/tex]The area to the left of z=-112.65 is 0.The area to the right of z=-112.65 is 1-0=1