"scores on a test have a mean of 63 and q3 is 80." the scores have a distribution that is approximately normal. find the standard deviation. round your answer to the nearest tenth.

Let X be the score on a test. X follows normal distribution with mean 63.

The third quartile q3 = 80

The third quartile is the value that separates data so that 3/4 = 0.75 area lies below q3 and 1/4 =0.25 area lies above q3

Now we know that q3 =80.

P(x < 80) = 0.75

Using z score table we will find z value corresponding to probability 0.75

As there is nor exact probability value 0.75 in table we will consider most closest value which is 0.7486. The z score corresponding to 0.7486 is 0.67

Now we will use z=0.67, mean =63 and x=80 to find standard deviation

0.67 = 17 / standard deviation

standard deviation = 17/0.67 = 25.37

The population standard deviation is 25.37

since y varies directly with y then

y = kx ( k is the constant of variation )

to find k use y = 3 when x = 2

k = =

equation is : y = x

When x = 1 then y = × 1 =