Inverse normal and normal practice Problem set A One question at a time

The height of a variety of sunflower is normally distributed with mean 210 cm and standard deviation 18 cm. The shortest 15% of plants are given extra fertiliser. Find the height below which a plant is given extra fertiliser.

• P(X < k) = 0.15
• From the calculator: inverse normal with area = 0.15, μ = 210, σ = 18
• k = 191

Scores on an aptitude test are normally distributed with mean 100 and standard deviation 15. The top 8% of scores are invited to an interview. Find the lowest score that gets an interview.

• P(X > k) = 0.08, so P(X < k) = 0.92
• From the calculator: inverse normal with area = 0.92, μ = 100, σ = 15
• k = 121

\(X \sim N(120, 15^2)\). Find the value of k such that \(P(110 < X < k) = 0.35\).

• From the calculator: normal cdf with lower = −10^99, upper = 110, μ = 120, σ = 15: 0.2525
• \(P(X < k) = 0.2525 + 0.35 = 0.6025\)
• From the calculator: inverse normal with area = 0.6025, μ = 120, σ = 15
• k = 124

The time taken by athletes to complete a 5 km run is normally distributed with mean 26.0 minutes and standard deviation 3.2 minutes.

1. (a) Find the lower quartile.

   • From the calculator: inverse normal with area = 0.25, μ = 26, σ = 3.2
   • Q1 = 23.8

2. (b) Find the upper quartile.

   • From the calculator: inverse normal with area = 0.75, μ = 26, σ = 3.2
   • Q3 = 28.2

3. (c) Find the interquartile range.

   • IQR = Q3 − Q1 = 4.32 (from the unrounded quartiles)

The daily sales of a café are normally distributed with mean 180 cups and standard deviation 25 cups. Find the probability that the café sells more than 200 cups on a day.

• \(X \sim N(180, 25^2)\)
• From the calculator: normal cdf with lower = 200, upper = 10^99, μ = 180, σ = 25
• P = 0.212

The mass of a strawberry is normally distributed with mean 24 g and standard deviation 3.5 g. In a delivery of 800 strawberries, find the expected number heavier than 30 g.

• From the calculator: normal cdf with lower = 30, upper = 10^99, μ = 24, σ = 3.5: 0.04324
• \(800 \times 0.04324 = 34.6\)
• About 35 strawberries

The time, in seconds, that a swimmer takes to swim 50 m is normally distributed with mean 31.4 and standard deviation 1.2. A coach calls the middle 60% of her times typical. Find the shortest and longest typical time.

• The middle 60% leaves 40%, so 20% in each tail
• From the calculator: inverse normal with area = 0.2, μ = 31.4, σ = 1.2: 30.4
• From the calculator: inverse normal with area = 0.8, μ = 31.4, σ = 1.2: 32.4
• The interval is 30.4 to 32.4 s

\(X \sim N(\mu, 6^2)\) and \(P(X < 40) = 0.3\). Find the value of \(\mu\).

• The unknown is μ. σ = 6
• From the calculator: solve normal cdf with lower = −10^99, upper = 40, μ = m, σ = 6 equal to 0.3, for m
• μ = 43.1

The time a player takes to complete a puzzle is normally distributed with mean 38 minutes and standard deviation 7 minutes. 90% of players finish within t minutes. Find t.

• \(P(X < t) = 0.9\)
• From the calculator: inverse normal with area = 0.9, μ = 38, σ = 7
• t = 47.0 minutes

The mass of a tomato is normally distributed with mean 70 g and standard deviation 12 g. A grower sorts 2 000 tomatoes. Small tomatoes (under 55 g) sell for €0.20 each, medium tomatoes (55 g to 80 g) for €0.35 each and large tomatoes (over 80 g) for €0.50 each. Find the expected income.

• P(small) = 0.106, P(medium) = 0.692, P(large) = 0.202 (calculator values, kept unrounded)
• Expected numbers: 211 small, 1384 medium, 405 large
• \(211 \times 0.20 + 1380 \times 0.35 + 405 \times 0.50\), using the unrounded numbers
• Expected income = €729.00

The time, in minutes, that a customer waits for a call to be answered is modelled by a normal distribution with mean 6 and standard deviation 3.5.

1. (a) Find the probability that the model gives a negative value.

   • From the calculator: normal cdf with lower = −10^99, upper = 0, μ = 6, σ = 3.5
   • P(X < 0) = 0.0432

2. (b) What does your answer tell you about the model?

   • A waiting time cannot be negative, but the model gives it probability 0.0432.
   • The mean is close to 0 relative to the spread, so the normal curve is a poor fit.

The diameter of a drill bit is normally distributed with mean 8.00 mm and standard deviation 0.05 mm. A drill bit is defective if its diameter is less than 7.9 mm. A batch of 10 bits is taken.

1. (a) Find the probability that one is below the limit given in the question.

   • From the calculator: normal cdf with lower = −10^99, upper = 7.9, μ = 8, σ = 0.05
   • p = 0.0228

2. (b) Find the probability that at least one bit in the batch is defective.

   • \(Y \sim B(10, 0.02275)\)
   • From the calculator: binomial cdf with n = 10, p = 0.02275, upper = 0: 0.7944
   • P(Y ≥ 1) = 1 − 0.7944 = 0.206

3. (c) State two assumptions you made in part (b).

   • The events are independent of each other.
   • The probability p is constant for every one.

A machine pours coffee into cups. The volume is normally distributed with mean 180 ml. 3% of cups contain less than 170 ml.

1. (a) Find the standard deviation.

   • X ~ N(180, σ²). P(X < 170) = 0.03
   • From the calculator: solve normal cdf with lower = −10^99, upper = 170, μ = 180, σ = s equal to 0.03, for s
   • σ = 5.32

2. (b) Find the probability that a cup holds more than 188 ml.

   • P(X > 188) = 1 − P(X < 188)
   • From the calculator: normal cdf with lower = 188, upper = 10^99, μ = 180, σ = 5.3169…: 0.0662

The masses of 400 rams are normally distributed with mean 75 kg and standard deviation 6 kg. The lightest ram has mass 60 kg and the heaviest 91 kg. Complete a box plot of the masses.

1. (a) Find the median, lower quartile and upper quartile.

   • The median of a normal distribution is the mean: 75
   • From the calculator: inverse normal with area = 0.25, μ = 75, σ = 6: 71.0
   • From the calculator: inverse normal with area = 0.75, μ = 75, σ = 6: 79.0

2. (b) Write the five numbers for the box plot and describe its shape.

   • Five numbers: 60, 71.0, 75, 79.0, 91
   • Draw the box from Q1 to Q3, the median line at the mean, and whiskers to the lowest and highest values. The plot is symmetric.

The time in minutes that a customer spends in a shop is modelled by a normal distribution with mean 5.2 and standard deviation 3.1.

1. (a) How many standard deviations is the mean above 0?

   • \(5.2 \div 3.1 = 1.68\)

2. (b) Explain why a normal distribution is not a suitable model.

   • 1.68 is less than 2.
   • From the calculator: normal cdf with lower = −10^99, upper = 0, μ = 5.2, σ = 3.1: 0.0467
   • The model gives a probability of 0.0467 to a negative time, which cannot happen.

The height, H metres, of the trees in a plantation is normally distributed. 12% of trees are shorter than 9 m and 30% are taller than 14 m.

1. (a) Find the mean and the standard deviation.

   • P(H < 9) = 0.12: inverse normal with area = 0.12, μ = 0, σ = 1: z = -1.1750
   • P(H > 14) = 0.3, so area = 0.7: z = 0.5244
   • μ − 1.1750σ = 9 and μ + 0.5244σ = 14
   • From the calculator: simultaneous equations: σ = 2.94, μ = 12.5

2. (b) Find the probability that a tree is more than 16 m.

   • From the calculator: normal cdf with lower = 16, upper = 10^99, μ = 12.4571…, σ = 2.9422…: 0.114