Answer:
(a) P(1000 < X < 1500) = 0.0256
(b) P(X < 1025) = 0.0392
(c) P(X > 1200) = 0.6554
(d) Percentile rank of a bag that contains 1425 chocolate chips = 90.98%
Step-by-step explanation:
We are given that the number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally distributed with a mean of 1252 chips and standard deviation 129 chips.
Firstly, Let X = number of chocolate chips in a bag
The z score probability distribution for is given by;
Z = 
~ N(0,1)
where, 
= population mean = 1252 chips
= standard deviation = 129 chips
(a) Probability that a randomly selected bag contains between 1000 and 1500 chocolate chips, inclusive is given by = P(1000 
X 1500) = P(X 1500) - P(X < 1000)
P(X 1500) = P( 
) = P(Z 1.92) = 0.9726
P(X < 1000) = P( < 
) = P(Z < -1.95) = 1 - P(Z 1.95)
= 1 - 0.9744 = 0.0256
Therefore, P(1000 X 1500) = 0.9726 - 0.0256 = 0.947
(b) Probability that a randomly selected bag contains fewer than 1025 chocolate chips is given by = P(X < 1025)
P(X < 1025) = P( < 
) = P(Z < -1.76) = 1 - P(Z 1.76)
= 1 - 0.9608 = 0.0392
(c) Proportion of bags contains more than 1200 chocolate chips is given by = P(X > 1200)
P(X > 1025) = P( > 
) = P(Z > -0.40) = P(Z < 0.40) = 0.6554
(d) <em>Percentile rank of a bag that contains 1425 chocolate chips is given by;</em>
Firstly we will calculate the z score of 1425 chocolate chips, i.e.;
Z = 
= 1.34
Now, we will check the area probability in z table which corresponds to this critical value of x;
The value which we get is 0.9098.
Therefore, 90.98% is the rank of bag that contains 1425 chocolate chips.
