The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed...

Here, we are given the distribution as:

Solution :

mean = = 1261

standard deviation = = 117

a) From standard normal tables we get:

P(Z < -0.6745) = 0.25

Therefore the 25th percentile observation here is computed as:

= Mean -0.6745*Std Dev.

= 1261 -0.6745*117

= 1182.08

b) From standard normal tables, we get:

P( -2.05 < Z <2.05) = 0.96

Therefore the interval here is computed as:

Mean - 2.05*Std Dev. , Mean+ 2.05*Std Dev.

1261 - 2.05*117, 1261 + 2.05*117

1020.71,1501.29

This is the required middle interval required here

c) From the standard normal tables, we get:

P(Z < -0.6745 ) = 0.25

Therefore, P(Z > 0.6745) = 0.25

Therefore, here the interval is computed as:

1261 - 0.6745*117, 1261 + 0.6745*117

Therefore interquartile range here is computed as:

= 2*0.6745*117

= 157.833

Therefore the required interquartile range is 157.833