Standard Normal Distribution

I can’t understand how we got -2.05 as our z-score in c.

Question:

The weights of a certain mass-produced item are known, over a long period of time, to be normally distributed with a mean of 8 kg and a standard deviation of 0.02 kg.

Required:

(a) If items whose weight lies outside the range 7.985 – 8.035 kg are deemed to be faulty, what percentage of products will be faulty?

(b) If it is required to reduce the percentage of items that are too heavy (with weight over 8.035 kg) to 2%, to what value must the mean weight be decreased, leaving all other factors unchanged?

(c) If it is required to reduce the percentage of items that are too light (with weight below 7.985 kg) to 2%, to what value must the standard deviation be decreased, leaving other factors unchanged?

Answer:

(a) 26.67% of products will be faulty If weight is 7.985 kg: z = x –? ? Z = (7.985 – 8) ÷ 0.02 = –0.75 If weight is 8.035 kg: Z = (8.035 – 8) ÷ 0.02 = 1.75 So we want P(–0.75 < z < 1.75) = TE(0.75) + TE(1.75) = 0.2734 + 0.4599 = 0.7333 Faulty items are outside this range so the probability that an item is faulty is 1 – 0.7333 = 0.2667.

(b) The mean weight must be decreased to 7.994 kg. The tail-end probability of 2% corresponds to the table entry 48% = 0.48, and so to the z-value 2.05. Use the formula to work back to calculate the mean: 2.05 = (8.035 – ?)/0.02 ? = 8.035 – (2.05 × 0.02) = 7.994.

(c) The standard deviation must fall to 0.0073 kg. From the logic above, the z-value must now be –2.05 Use the formula to work back to calculate the standard deviation: –2.05 = (7.985 – 8)/? ? = –0.015 ÷ –2.05 = 0.0073