Question

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.

(a) what is the probability that a randomly selected bag contains between 1000 and 1500 chocolate chips, inclusive?

(b) what is the probability that a randomly selected bag contains fewer than 1125 chocolate chips?

(c) what proportion of bags contains more than 1175 chocolate chips?

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

(a) the probability that a randomly selected bag contains between 1000 and 1500 chocolate chips, inclusive is _. (round to four decimal places as needed.)

(b) the probability that a randomly selected bag contains fewer than 1125 chocolate chips is _. (round to four decimal places as needed.)

(c) the proportion of bags that contains more than 1175 chocolate chips is _. (round to four decimal places as needed.)

(d) a bag that contains 1475 chocolate chips is in the _th percentile. (round to the nearest integer as needed.)

Answer 1

Solution :

Given that ,

mean = = 1252

standard deviation = = 129

(a)

P(1000 < x < 1500) = P((1000 - 1252)/ 129) < (x - ) /  < (1500 - 1252) / 129) )

= P(-1.95 < z < 1.92)

= P(z < 1.92) - P(z < -1.95)

= 0.9726 - 0.0256

= 0.947

Probability = 0.947

(b)

P(x < 1125) = P((x - ) / < (1125 - 1252) / 129)

= P(z < -0.98)

= 0.1635

Probability = 0.1635

(c)

P(x > 1175) = 1 - P(x < 1175)

= 1 - P((x - ) / < (1175 - 1252) / 129)

= 1 - P(z < -0.60)

= 1 - 0.2743   

= 0.7257

Probability = 0.7257

(d)

(x - ) / = (1475 - 1252) / 129 = 1.73

P(z < 1.73) = 0.96

96th percentile