Question

In a large human population, cranial length is approximately normally distributed with a mean of 185.6 mm and a standard deviation of 12.7 mm. What is the probability that a random sample of size 10 from this population will have a mean less than than 180? a. 0.0174 b. 0.1357 c. 0.0823 d. -1.39

Answer 1

The correct option is c

Explanation:

From the question we are told that

The mean is
`\mu = 185.6 \ mm`

The standard deviation is
`\sigma = 12.7 \ mm`

The sample size is n = 10

Generally the standard error of the mean is mathematically represented as

`\sigma_(x) = (\sigma)/(√(n) )`

=>
`\sigma_(x) = ( 12.7)/(√( 10 ) )`

=>
`\sigma_(x) = 4.016`

Generally the probability that a random sample of size 10 from this population will have a mean less than than 180 is mathematically represented as

`P(X < 180 ) = P(( X - \mu )/(\sigma_(x) ) < (180 - 185.6)/(4.016) )`

`(X -\mu)/(\sigma )  =  Z (The  \ standardized \  value\  of  \ X )`

`P(X < 180 ) = P(Z < - 1.394 )`

From the z table the area under the normal curve to the left corresponding to -1.394 is

`P(X < 180 ) = P(Z < - 1.394 ) = 0.081659`

=>
`P(X < 180 ) = 0.0823`