Question

A population of men is monitoring their daily caloric intake for training purposes. They find that their caloric intake is approximately normally distributed with a population mean of 2500 calories and a population standard deviation of 500 calories.

1. Find the probability that a randomly selected male from this population has a caloric intake of between 1500 and 3500.

2. Find the probability that a randomly selected male from this population has a caloric intake less than 1500.

3. Find the 93rd percentile of the caloric intake of this population of men.

Answer 1

Solution :

Given that ,

mean = = 2500

standard deviation = = 500

P(1500< x <3500 ) = P[(1500-2500) /500 < (x - ) / < (3500-2500) /500 )]

= P( -2< Z <2 )

= P(Z <2 ) - P(Z <-2 )

Using z table   

= 0.9772-0.0228

probability= 0.9544

(B)

P(X<1500 ) = P[(X- ) / < (1500-2500) /500 ]

= P(z < -2)

Using z table

= 0.0228

probability=0.0228

(C)

Using standard normal table,

P(Z < z) = 93%

= P(Z < z) = 0.93  

= P(Z <1.48 ) = 0.93

z = 1.48 Using standard normal table,

Using z-score formula  

x= z * +

x=1.48 *500+2500

x= 3240