Question

High school girls average 80 text messages daily. Assume the population standard deviation is 15 text messages. Assume normality.

1. What is the probability that the sample mean is more than 90?
2. What is the probability that the sample mean is less than 85?
3. What is the probability the sample mean is in between 85 and 90?

Answer 1

Solution :

Given that,

mean = = 80

standard deviation = = 15

a ) P (x > 90 )

= 1 - P (x < 90 )

= 1 - P ( x -  / ) < ( 90 - 80 / 15)

= 1 - P ( z <10 / 15 )

= 1 - P ( z < 0.67)

Using z table

= 1 - 0.7486

=0.2514

Probability = 0.2514

b ) P( x < 85 )

P ( x - / ) < ( 85 - 80 / 15)

P ( z < 5 / 15 )

P ( z < 0.33)

=0.6293

Probability = 0.6293

c ) P (85 < x < 90 )

P ( 85 - 80 / 15) < ( x -  / ) < ( 90 - 80 / 15)

P (5 / 15 < z < 10 /15 )

P (0.33 < z < 0.67)

P ( z < 0.67 ) - P ( z < 0.33)

Using z table

= 0.7486 - 0.6293

= 0.1193

Probability =0.1193