Hint: Convert the normal distribution X to Standard normal using Z formula Z=(X-μ)/σZ-values from the table and then find the probability.

Question

A survey indicates that for each trip to a supermarket, a shopper spends an average of 43 minutes with a standard deviation of 12 minutes in the store. The lengths of time spent in the store are normally distributed and are represented by the variable X. A shopper enters the store. Find the probability that the shopper will be in the store for each interval of time listed below.

a) Find the probability that the shopper will be in the store between 33 and 66 minutes.

b) Find the probability that the shopper will be in the store for more than 39 minutes.

Hint: Convert the normal distribution X to Standard normal using Z formula Z=(X-μ)/σ and then look the Z-values from the table and then find the probability.

Hint: Convert the normal distribution X to Standard normal using Z formula Z=(X-μ)/σ and then look the Z-values from the table and then find the probability.

Answer 1

Remember that the z-table show probabilities (area under the Normal Distribution Curve) to the left of the z-value. When finding the probability of a range you'll need to find the two relevant z-factors and subtract them to find the area (probability) for the given range. So here goes:

1) P[33<x<66] For 33 min., z = (x - μ)/σ = (33 - 43)/12 = -10/12 = -0.8333_. On the z- table this equates to a probability of spending less than min. of 0.2033 (I used -0.83 , close enough!) For 66 min. z =(63 - 43)/12 = 1.9167. Using 1.92, the probability of less than 66 min. is 0.9726.

So the answer to (1) is 0.9726 - 0.2033, or 0.7693, or about 77%.

2) For the probability of spending more than 39 min, need first to find P[x<39]. z = (39 - 43)/12 = -0.333_.

The z-value is 0.3707. But that's the probability of fewer than 39 min. For probabiity of more than 39 min. P[x>39], do 1 - 0.3707 = 0.6293, or about 63%.

Answer 2

The trick to this problem is to "standardize" the random variable X by subtracting its mean from it and then dividing the result by its standard deviation. After doing this to any Normally distributed random variable, the result will have a Standard Normal distribution.

In our case, the variable X has distribution N(43,12). Its mean is 43 and its standard deviation is 12. Therefore the random variable

X - 43

has distribution N(0,12), and the random variable

(X-43) / 12

has distribution N(0,1) - it has Standard Normal distribution. This means we can look up the probability that it falls in a particular range of numbers by using a Standard Normal Distribution table. Let's apply this in our case.

(a) We want to know "the probability that X is between 33 and 66", or

P[33 < X < 66].

Another way of saying "X is between 33 and 66" is to say "(X minus 43) is between (33 minus 43) and (66 minus 43)". So the probability we're interested in finding can be equivalently expressed as

= P[33-43 < X-43 < 66-43]

or

= P[-10 < X-43 < 23].

Similarly, saying "(X minus 43) is between -10 and 23" is equivalent to saying "(X minus 43)/12 is between -10/12 and 23/12". So this probability can be rewritten as

= P[-10/12 < (X-43)/12 < 23/12].

From here, we use the fact that (X-43)/12 has Standard Normal distribution. To emphasize this fact, let's call (X-43)/12 by the name "Z", a name usually used to denote random variables with Standard Normal distribution. Converting the fractions to decimals, this means we can rewrite the above as

= P[-0.833 < Z < 1.917].

Now we're almost ready to use the Standard Normal table to find the answer! Before we can, though, we should remember that most Standard Normal tables only allow you to look up probabilities written in the form "P[Z < x]" for some number x. For example, we could use a Standard Normal table to find P[Z < 1.917] or to find P[Z < -0.833], but we can't (directly) use it to find P[Z > -0.833] (since this has a ">" instead of a "<"). In order to find P[-0.833 < Z < 1.917], we'll have to use a trick. We have to rewrite it as

= P[Z < 1.917] - P[Z < -0.833].

Why is this a correct way to rewrite P[-0.833 < Z < 1.917]? Well, P[-0.833 < Z < 1.917] equals the area under the Standard Normal density curve between the x-values -0.833 and 1.917. Another way of computing this area is to first find the area under the curve between negative infinity and 1.917 and then subtract from this the area under the curve to between negative infinity and -0.833. Mathematically, the first area is P[Z < 1.917] and the second area we're subtracting is P[Z < -0.833]. So the expression P[Z < 1.917] - P[Z < -0.833] is exactly equal to P[-0.833 < Z < 1.917].

Now looking numbers up in the Standard Normal table, we find P[Z < 1.917] = .972 and P[Z < -0.833] = .202. So our answer is

= .972 - .202 = .770,

or about a 77% chance.

(b) We want to find

P[X > 39].

Using similar steps to the above, we can rewrite this as

=P[X-43 > 39-43]

=P[X-43 > -4]

=P[(X-43)/12 > -4/12]

=P[Z > -0.333].

Again, we can't directly look up P[Z > -0.333] since this has a ">" instead of a "<". But remembering that the total area under the Standard Normal density curve equals 1, we can use a trick. We know that

(Area under curve between negative infinity and -0.333) + (Area under curve between -0.333 and positive infinity) = 1,

so in particular

(Area under curve between -0.333 and positive infinity) = 1 - (Area under curve between negative infinity and -0.333).

Mathematically, this says

P[Z > -0.333] = 1 - P[Z < -0.333].

We can look up P[Z < -0.333] in a Standard Normal table; it equals .370. So we have

P[Z > -0.333] = 1 - .370 = .630,

or about a 63% chance.