Question

The number of chocolate chips in an​ 18-ounce bag of chocolate chip cookies is approximately normally distributed with mean 1252 and standard deviation 129 chips. ​

(a) What is the probability that a randomly selected bag contains between 1000 and 1400 chocolate​ chips? ​

(b) What is the probability that a randomly selected bag contains fewer than 1100 chocolate​ chips? ​

(c) What proportion of bags contains more than 1225 chocolate​ chips?

​(d) What is the percentile rank of a bag that contains 1475 chocolate​ chips?

Answer 1

Solution :

Given that ,

mean = = 1252

standard deviation = = 129

P(1000< x <1400 ) = P[(1000-1252) /129 < (x - ) / < (1400-1252) /129 )]

= P( -1.95< Z <1.15 )

= P(Z <1.15 ) - P(Z <-1.95 )

Using z table   

=  0.8749-0.0256

probability= 0.8493

B)P(X<1100 ) = P[(X- ) / < (1100-1252) / 129]

= P(z <-1.18 )

Using z table

=0.1190

probability=0.1190

(c)P(x >1225 ) = 1 - P(x<1225 )

= 1 - P[(x -) / < (1225-1252) / 129]

= 1 - P(z <-0.21 )

Using z table

= 1 -  0.4168

=0.5832

prportion=0.5832

(e)P(X<1475 ) = P[(X- ) / < (1475-1252) / 129]

= P(z <1.73 )

Using z table

=0.9582

percentile rank=95.82% rounded=96%