Question

ThenumberofdailytextssentbyMarymountstudentsarenormally distributed with a mean of 80 texts and a standard deviation of 50 texts.

(a) Find the probability that a randomly selected Marymount student sends more than 100 texts each day.

(b) Find the probability that 25 randomly selected Marymount students will have a mean number of daily texts sent that is greater than 50 texts.

(c) Suppose a parent wants their child in the bottom 25% of texters. Find the cut-off value for the number of texts below which 25% of MCU students lie.

Answer 1

Solution :

Given that,

mean = = 80

standard deviation = = 50

P(x > 100) = 1 - P(x<100 )

= 1 - P[(x -) / < (100-80) /50 ]

= 1 - P(z < 0.4)

Using z table

= 1 -  0.6554

probability= 0.3446

(B)

n=25

= =80

= / n = 50/ 25= 10

P( >50 ) = 1 - P( <50 )

= 1 - P[( - ) / < (50-100) /10 ]

= 1 - P(z <-5 )

Using z table

= 1 - 0

= 1

probability= 1

(C)

using standard normal table

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.67 ) = 0.25

z = -0.67

Using z-score formula,

x = z * +

x = -0.67 * 50+100

x = 66.5