The population distribution of SAT scores is normal with a mean of μ=500 and a standard deviation of SD=100. For example, what is the probability of randomly selecting an individual from this population who has an SAT score greater than 700?

0.02275

Step-by-step explanation:

We use the z score formula to solve for this

z-score is given as: z = (x-μ)/σ

where x is the raw score,

μ is the population mean

σ is the population standard deviation

In the above question:

mean of μ=500

a standard deviation of SD=100

raw score x = 700

Hence, z score = (700 - 500)/ 100

= 200/100

= 2

z score = 2

Using the z score table of normal distribution to find the Probability of z = 2

P( x = z)

= P(x = 700)

= P( z = 2)

= 0.97725

P(x>700) = 1 - P(x = 700)

= 1 - 0.97725

= 0.02275

Therefore, the probability of randomly selecting an individual from this population who has an SAT score greater than 700 is 0.02275

it is contrapositive

i could not see the question too well so check yourself again too . this is because i want you to get it correct.

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)