Question

1. A distribution of values is normal with a mean of 70.8 and a standard deviation of 50.9.

Find the probability that a randomly selected value is less than 4.6.
P(X < 4.6) =

2. A distribution of values is normal with a mean of 66 and a standard deviation of 4.2.

Find the probability that a randomly selected value is greater than 69.4.
P(X > 69.4) =

Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

Solution :

Given that,

mean = =70.8

standard deviation = = 50.9

P(X<4.6 ) = P[(X- ) / < (4.6-70.8) /50.9 ]

= P(z <-1.301 )

Using z table

=0.0966

b.

Solution :

Given that,

mean = = 66

standard deviation = = 4.2

P(x >69.4 ) = 1 - P(x< 69.4)

= 1 - P[(x -) / < (69.4-66) /4.2 ]

= 1 - P(z <0.810)

Using z table

= 1 -  0.7910

probability= 0.2090