# Video: EG19M1-Statistics-Q7B

The monthly salary of 1000 families in a city is a normal random variable whose mean is 170 LE and standard deviation is 20 LE. Find the number of families whose salaries are more than 150 LE.

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### Video Transcript

The monthly salary of 1000 families in a city is a normal random variable whose mean is 170 Egyptian pounds and standard deviation is 20 Egyptian pounds. Find the number of families whose salaries are more than 150 Egyptian pounds.

In general, a normal random variable can be expressed in terms of its mean 𝜇 and standard deviation 𝜎. In this question, if we let 𝑋 be the distribution of monthly salaries, we have that 𝑋 follows a normal distribution with mean 170 and standard deviation 20. which means the variance of the distribution is 20 squared. Before we can work out the number of families whose salaries are more than 150 Egyptian pounds, we must first work out the probability that a randomly chosen family from this distribution has a salary of more than 150 pounds.

We begin by recalling that the normal distribution is a bell-shaped curve symmetrical about its mean, which in this case is 170. The area underneath the full curve is equal to one. And the area to the left of any particular value gives the probability that a randomly chosen value from this distribution is less than or equal to that value. We however want the probability that 𝑋 is greater than a particular value. We want the probability that 𝑋 is greater than 150, which means we want to work out the area to the right of 150.

Remember that the full area under the curve is equal to one. So these two areas sum to one. To find probabilities for a normal distribution, we must first work out the 𝑍-score associated with a particular value. The 𝑍-score tells us how many standard deviations above or below the mean a particular value is. And we find it by taking the value 𝑥 subtracting the mean 𝜇 and then dividing by the standard deviation 𝜎. In this question, our value 𝑥 is 150, the mean 𝜇 is 170, and the standard deviation 𝜎 is 20. So we have 150 minus 170 over 20. This simplifies to negative 20 over 20 which is just equal to negative one.

The interpretation of this value of negative one is that 150 is one standard deviation below the mean, which makes sense. 150 is 20 less than 170 and 20 is the standard deviation of this distribution. We now use our standard normal distribution tables to look up the probability associated with a 𝑍-score of negative one. We find negative one on the left-hand side of our tables. And as our 𝑍-score is exactly negative one, so negative 1.00, we’re looking in the first column of the table. We see that the probability associated with the 𝑍-score of negative one is 0.1587. This tells us that the probability of a 𝑍 score less than or equal to negative one is 0.1587. And in turn, the probability that our normal random variable 𝑋 is less than or equal to 150 is also 0.1587.

Remember though that we want to know the probability that 𝑋 is greater than 150 so we can work out the number of families whose salaries are more than 150 Egyptian pounds. So using the fact that the full area below the curve is one, we can subtract the value of 0.1587 from one to give 0.8413. So now, we know that the probability of getting an 𝑋-value from this distribution greater than 150 is 0.8413 which is the probability that a randomly chosen family will have a salary more than 150 Egyptian pounds.

To work out the number of families that have a salary more than 150 Egyptian pounds, we use the fact that there are 1000 families living in this city. We take the number of families 1000 and multiply it by the probability of each family having a salary more than 150 Egyptian pounds. That’s 1000 multiplied by 0.8413. That gives 841.3. But we need the number of families to be an integer. So we’ll round our value to the nearest whole number. That’s 841.

We found then that the number of families in this city whose salaries are more than 150 Egyptian pounds is 841 correct to the nearest integer.