Potatoes: Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of...

Question:

Potatoes: Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 8.5 ounces and a standard deviation of 1.1 ounces. Round your answers to 4 decimal places.

(a) If one potato is randomly selected, find the probability that it weighs less than 7 ounces.

(b) If one potato is randomly selected, find the probability that it weighs more than 10 ounces.

(c) If one potato is randomly selected, find the probability that it weighs between 7 and 10 ounces.

Normal Probability Problem:

The formula of z score is used to get the probabilities of weights. The z score is a normal score and it is also known as the standard normal variate with mean zero and variance one. The square of standard normal variate is known as the chi-square variate with 1 degree of freedom. An excel function of normal distribution NORM.DIST() is also used to get the required probabilities.

Answer and Explanation:

Given that,

  • Mean, {eq}\mu = 8.5 {/eq}
  • Standard deviation, {eq}\sigma = 1.1 {/eq}


a)

The required probability is {eq}P(X < 7). {/eq}


Now,

{eq}P(X < 7) = P(\dfrac{X - \mu}{\sigma} < \dfrac{7 - \mu}{\sigma})\\ P(X < 7) = P(Z < \dfrac{7 - 8.5}{1.1})\\ P(X < 7) = P(Z < -1.363) {/eq}


Excel function for the above probability:

=NORM.DIST(-1.363,0,1,1)


{eq}P(X < 7) = 0.0864 {/eq}

The probability that it weighs less than 7 ounces = 0.0864.


b)

The required probability is {eq}P(X > 10). {/eq}


Now,

{eq}P(X > 10) = 1 - P(\dfrac{X - \mu}{\sigma} \le \dfrac{10 - \mu}{\sigma})\\ P(X > 10) = 1- P(Z \le \dfrac{10 - 8.5}{1.1})\\ P(X > 10) = 1 - P(Z \le 1.363) {/eq}


Excel function for the above probability:

=NORM.DIST(1.363,0,1,1)


{eq}P(X > 10) = 1 - 0.91356\\ P(X > 10) = 0.0864 {/eq}

The probability that it weighs more than 10 ounces = 0.0864.


c)

The required probability is {eq}P(7 < X < 10). {/eq}


Now,

{eq}P(7 < X < 10) = P(\dfrac{7 - \mu}{\sigma} < \dfrac{X - \mu}{\sigma} < \dfrac{10 - \mu}{\sigma})\\ P(7 < X < 10) = P( \dfrac{7 - 8.5}{1.1} < Z < \dfrac{10 - 8.5}{1.1}\\ P(7 < X < 10) =P(-1.363 < Z < 1.363)\\ P(7 < X < 10) =P(Z < 1.363) - P(Z < -1.363) {/eq}


Excel function for the above probability:

=NORM.DIST(1.363,0,1,1)

=NORM.DIST(-1.363,0,1,1)


{eq}P(7 < X < 10) = 0.9135 - 0.0864 P(7 < X < 10) = 0.8271 {/eq}

The probability that it weighs between 7 and 10 ounces = 0.8271.


Learn more about this topic:

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Using the Normal Distribution: Practice Problems

from Statistics 101: Principles of Statistics

Chapter 6 / Lesson 9
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