Finding Probability Given a Basic Word Problem Involving Normal Distribution

Finding Probability Given a Basic Word Problem Involving Normal Distribution

Step 1: Write the probability desired in probability notation in terms of the raw score, x. Note both the mean, {eq}\mu {/eq}, and standard deviation of the distribution of raw scores.

Step 2: Standardized the raw score, x, using the formula {eq}z=\frac{x-\mu}{\sigma} {/eq} to get the standardized score and we rewrite the probability statement in terms of z.

Step 3: Find the probability on a standard normal distribution table or with technology. This is the answer.

Finding Probability Given a Basic Word Problem Involving Normal Distribution Vocabulary and Formula

Raw Score: The raw score is the endpoint in the original probability statement. It is in the same units as the original data.

Mean: The mathematical average of all possible values of x. It is represented by the symbol {eq}\mu {/eq}.

Standard Deviation: The standard deviation is the average distance, or deviation, each possible observation is from the mean. It is represented by {eq}\sigma {/eq}.

Standardized Score: The standardized score, z, of a raw score is the number of standard deviations above or below the mean that the raw score lies. It is calculated with the formula: {eq}\frac{x-\mu}{\sigma} {/eq}.

Standard Normal Table: A standard normal table is a table giving probabilities associated with standardized scores, the z scores. It has a mean of 0 and a standard deviation of 1.

Now, we are ready to consider two examples, one with a less than in the probability statement and one with a greater than in the probability statement.

Finding Probability Given a Basic Word Problem Involving Normal Distribution: Less Than Example

The grades on a test are normally distributed with a mean of 82 and a standard deviation of 10. What is the probability that a randomly selected grade is below 75?

Step 1: First, we write the probability desired in probability notation in terms of the raw score, x. Note both the mean, {eq}\mu {/eq}, and standard deviation,{eq}\sigma {/eq}, of the distribution of raw scores.

P(x<75) since we want the probability a randomly selected grade is below 75, which is our raw score.

{eq}\mu=84 {/eq} since the mean is 84.

{eq}\sigma=10 {/eq} since the standard deviation is 10.

Step 2: Then, we standardized the raw score, x, using the formula {eq}z=\frac{x-\mu}{\sigma} {/eq} to get the standardized score and we rewrite the probability statement in terms of z.

{eq}z=\frac{x-\mu}{\sigma} {/eq}

{eq}z=\frac{75-85}{10}=\frac{-7}{10}=-0.70 {/eq}

{eq}P(X<75)= P(z<-0.70) {/eq}

Step 3: Finally, we find the probability on a standard normal distribution table or with technology. This is the answer.

{eq}P(z<-0.70)=0.2420 {/eq} when rounded to four decimal places. We can get this probability from a Standard Normal Table or by using technology.

The probability a randomly selected grade will be below 75 is 0.2420.

Finding Probability Given a Basic Word Problem Involving Normal Distribution: Greater Than Example

Crazy Crunchy Potato Chips are sold in bags with a mean weight of 10 ounces and a standard deviation of 0.2 ounces. What is the probability of getting a bag of chips that weighs more than 10.25 ounces?

Step 1: First, we write the probability desired in probability notation in terms of the raw score, x. Note both the mean, {eq}\mu {/eq}, and standard deviation, {eq}\sigma {/eq}, of the distribution of raw scores.

P(X>10.25) since we want the probability a bag of chips weighs more than 10.25 ounces.

{eq}\mu=10 {/eq}

{eq}\sigma=0.2 {/eq}

Step 2: Then, we standardized the raw score, x, using the formula {eq}z=\frac{x-\mu}{\sigma} {/eq} to get the standardized score and we rewrite the probability statement in terms of z.

{eq}z=\frac{x-\mu}{\sigma} {/eq}

{eq}z=\frac{10.25-10}{0.2} {/eq}

{eq}z=\frac{0.25}{0.2}=1.25 {/eq}

So our probability statement in terms of z is {eq}P(z>1.25) {/eq}

Step 3: Finally, we find the probability on a standard normal distribution table or with technology. This is the answer.

{eq}P(z>1.25)=0.1057 {/eq} when rounded to four decimal places. We can get this probability from a Standard Normal Table or by using technology.

The probability of getting a bag of Crazy Crunchy Potato Chips that weighs more than 10.25 ounces is 0.1057.