Question

The number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally distributed with a mean of 1252 chips and standard deviation of 129 chips.
a. What is the probability that a randomly selected bag contains between 1000 and 1400 chocolate chips?

b. What is the probability that a randomly selected bag contains more than 1225 chocolate chips?

c. What is the percentile rank of a bag that contains 1425 chocolate chips

Suppose the lengths of pregnancies of a certain animal are approximately normally distributed with mean ? = 240 days and ? = 18 days.

a. What is the probability that a randomly selected pregnancy lasts less than 233 days?

b. Suppose a random sample of 17 pregnancies is obtained. Describe the mean and standard deviation of the distribution of the sample mean length of pregnancies.

c. What is the probability that a random sample of 17 pregnancies has a mean pregnancy of less than 233 days?

According to a study conducted by an organization, the proportion of Americans who were afraid to fly in 2006 was 0.10. A random sample of 1200 Americans results in 108 indicating that they are afraid to fly.

a. What is the sampling distribution of the proportion of Americans who are afraid to fly?

b. What is the point estimate of the sample (?̂)?

c. What is the probability that 108 or fewer out of 1200 Americans are afraid to fly?

d. Is this evidence that the proportion of Americans who are afraid to fly is decreasing? Why or why not?

Please show all work for the following problems. If a calculator is used, include the calculator function used as well as the values entered. (Example: normalcdf

Answer 1

Solution :

1) Let X be a random variable which represents the number of chocolate chips in an 18-ounce bag of chocolate chip cookies.

Given that,

i.e. Mean (μ) = 1252 chips

Standard deviation (σ) = 129 chips

a) We have to obtain P(1000 < X < 1400).

We know that if X ~ N(μ, σ​​​​​​2) then

Using "pnorm" function of R we get,

P(Z < 1.1473) = 0.8744 and P(Z < -1.9535) = 0.0254

Hence, the probability that a randomly selected bag contains between 1000 and 1400 chocolate chips is 0.8490.

b) We have to obtain P(X > 1225).

We know that if X ~ N(μ, σ​​​​​​2) then

Using "pnorm" function of R we get, P(Z > -0.2093) = 0.5829

Hence, the probability that a randomly selected bag contains more than 1225 chocolate chips is 0.5829.

c) Let the percentile rank of a bag that contains 1425 chocolate chips is k.

Now if percentile rank of 1425 chocolate chips is k then,

P(X < 1425) = k.

So, basically we need to find P(X < 1425).

We know that if X ~ N(μ, σ​​​​​​2) then

Using "pnorm" function of R we get, P(Z < 1.3411) = 0.9101

0.9101 = 91.01%

k = 91.01%

Hence, the percentile rank of a bag that contains 1425 chocolate chips is 91.01th.