Solve the problem. The scores on a certain test are normally distributed with a mean score of 60 and a standard deviation of 5. What is the probability that a sample of 90 students will have a mean score of at least 60.527? Write your answer as a decimal rounded to 4 places.

15.87%

Step-by-step explanation:

We have to calculate the value of z:

z = (x - m) / (sd / n ^ (1/2))

where x is the value to evaluate, m is the mean, n is the sample size and sd is the standard deviation, we replace:

p (x <60,527) = z = (x - m) / (sd / n ^ (1/2))

p (x <60,527) = z = (60,527 - 60) / (5/90 ^ (1/2))

z = 1

if we look in the attached table, for z = 1 it is 0.8413

p (x> 60,527) = 1 - 0.8413

 p (x> 60,527) = 0.1587

Therefore the probability is 15.87%

d.192

step-by-step explanation:

12*16=192

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: