AI SL · 13.2
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Normal probability calculations with the GDC
13.2.8
Screenshot and video to come
Screenshot and video to come
Screenshot and video to come
A probability is an area under the curve. Drag the top of the curve to move the mean, the teal point on its side to change σ, and the orange points under the axis to choose the area. Or type exact values. Say how big the area is, then press Show answer.
Drag with the mouse or a finger, or select a point (Tab) and use the arrow keys. The axis fits itself to the curve when you let go.
Drag a point to change the shaded area.
The time a parcel takes to be delivered is normally distributed with mean 48 hours and standard deviation 9 hours. Find the probability that a parcel takes more than 60 hours.
Have a go first. Then open the working one line at a time.
Let \(X\) be the delivery time in hours. \(X \sim N(48, 9^2)\)
Write the distribution before you use the calculator.
From the calculator: normal cdf with lower = 60, upper = 10^99, μ = 48, σ = 9
More than 60 has no upper limit, so use a very large number such as 10^99.
\(P(X > 60) = 0.0912\)
About 9.1% of parcels take more than 60 hours.
The mass of flour in a bag is normally distributed with mean 1 000 g and standard deviation 12 g. Find the probability that a bag contains less than 985 g.
Have a go first. Then open the working one line at a time.
\(X \sim N(1 000, 12^2)\), \(X\) in grams
State the mean and the standard deviation.
From the calculator: normal cdf with lower = −10^99, upper = 985, μ = 1 000, σ = 12
Less than 985 has no lower limit, so use a very large negative number such as −10^99.
\(P(X < 985) = 0.106\)
For a continuous variable, P(X < 985) and P(X ≤ 985) are the same.
The length of a fish of one species is normally distributed with mean 32 cm and standard deviation 4.5 cm. Find the probability that a fish is between 28 cm and 36 cm long.
Have a go first. Then open the working one line at a time.
\(X \sim N(32, 4.5^2)\)
Mean 32 and standard deviation 4.5.
From the calculator: normal cdf with lower = 28, upper = 36, μ = 32, σ = 4.5
Put the smaller value in lower and the larger in upper.
\(P(28 < X < 36) = 0.626\)
The limits are symmetric about the mean, so this is also twice the area from 32 to 36.
The volume of juice in a carton, \(X\) ml, is modelled by \(X \sim N(250, 16)\). Find the probability that a carton contains less than 246 ml.
Have a go first. Then open the working one line at a time.
\(\sigma^2 = 16\), so \(\sigma = \sqrt{16} = 4\)
The second number in N(250, 16) is the variance. The calculator needs the standard deviation.
From the calculator: normal cdf with lower = −10^99, upper = 246, μ = 250, σ = 4
Enter σ = 4.
\(P(X < 246) = 0.159\)
The standard deviation is 4, not 16.
The marks in a test are normally distributed with mean 62 and standard deviation 8. Sketch the normal curve, shade the region for P(X > 70) and find the probability.
Have a go first. Then open the working one line at a time.
Bell curve centred at \(62\); mark \(62\) on the axis and \(70\) to the right.
Shade to the right of 70. The shaded area is the probability.
From the calculator: normal cdf with lower = 70, upper = 10^99, μ = 62, σ = 8
The shaded region has no upper limit.
\(P(X > 70) = 0.159\)
The shaded region is a small part of the curve, so a probability below 0.5 makes sense.
The box plot shows the number of minutes that 40 customers waited to be served. Is it reasonable to model the waiting time with a normal distribution?
Have a go first. Then open the working one line at a time.
Median − Q1 = 2 and Q3 − median = 6
A symmetric distribution has the two halves of the box about equal.
Lower whisker 2, upper whisker 18
The upper tail is much longer: skewed to the right.
Not reasonable.
A normal curve is symmetric. A skewed box plot means a normal model is poor.