Question

A quick survey of peanut butter prices had standard deviation and mean of $0.26 and $3.68,...

A quick survey of peanut butter prices had standard deviation and mean of $0.26 and $3.68, respectively. Compute the area for a peanut butter jar costing between $3.50 and $4.25.

Homework Answers

Answer #1

Solution :

Given that ,

mean =   = $3.68

standard deviation = = $0.26   

P($3.50< x <$4.25 ) = P[(3.50-3.68) /0.26 < (x - ) / < (4.25-3.68) /0.26 )]

= P( -0.69< Z < 2.19)

= P(Z <2.19 ) - P(Z <-0.69 )

Using z table   

= 0.9857-0.2451

probability= 0.7406

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
a JAR OF PEANUT BUTTER CONTAINS 451G WITH A STANDARD DEVIATION OF 10.2G. FIND THE PROBABILITY...
a JAR OF PEANUT BUTTER CONTAINS 451G WITH A STANDARD DEVIATION OF 10.2G. FIND THE PROBABILITY THAT A JAR CONTAINS MORE THAN 457G
A jar of peanut butter contains 460g with a standard deviation of 10.2g. Find the probability...
A jar of peanut butter contains 460g with a standard deviation of 10.2g. Find the probability that a jar contains less than 448g. Assume a normal distribution. The probability the jar contains less than 448 g is _____
A survey of 25 randomly selected customers had the mean 33.04 years and the standard deviation...
A survey of 25 randomly selected customers had the mean 33.04 years and the standard deviation 10.45 years. What is the standard error of the mean? How would the standard error change if the sample size had been             400 instead of 25? (Assume that the sample standard deviation didn't change.) Find the degrees of freedom for n=60. Find the t critical value for n=60 for 90% CI for mean. A survey of 25 randomly selected customers had the mean...
A survey found that? women's heights are normally distributed with mean 63.7 in and standard deviation...
A survey found that? women's heights are normally distributed with mean 63.7 in and standard deviation 2.2 in The survey also found that? men's heights are normally distributed with mean 67.5 in and standard deviation 3.9 in Consider an executive jet that seats six with a doorway height of 55.7 in Complete parts? (a) through? (c) below. a. What percentage of adult men can fit through the door without? bending? The percentage of men who can fit without bending is...
For a continuous random variable XX, the population mean and standard deviation are 118 and 11,...
For a continuous random variable XX, the population mean and standard deviation are 118 and 11, respectively. A sample of 42 observations is randomly selected. The mean of the sampling distribution of x¯ is: 118 1.18 1.7 11 For the standard normal distribution, the area between z=-1.58 and z=1.57is: 0.9418 0.9936 0.0014 0.8847 For the standard normal distribution, the area to the left of z=-1.06 is: 0.0082 0.1446 0.8554 0.0764
The mean and standard deviation of single sales prices for a specific type of golf club...
The mean and standard deviation of single sales prices for a specific type of golf club on Ebay are $147 and $11 respectively. We are going to sell 99 of these clubs on Ebay in individual and independent auctions. If the sales follow the historical data, what’s the maximum revenue we should expect, using a 98% CI as a basis? a. ~ $14,585 b. ~ $14,808 c. ~ $17,295 d. ~ $14,835 e. ~ $21,275
Suppose a variable has a normal distribution with mean 12 and standard deviation 2. Use the...
Suppose a variable has a normal distribution with mean 12 and standard deviation 2. Use the Empirical Rule to calculate the approximate PERCENTAGE area. What is the PERCENTAGE of values ABOVE 10? Note: Enter X.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 7 is entered as 7.00, 3.5 is entered as 3.50, 0.3750 is entered as 0.38 |Enter PERCENTAGE in above blank with NO % sign. | Suppose a variable has a normal distribution...
A survey found that​ women's heights are normally distributed with mean 63.6 in and standard deviation...
A survey found that​ women's heights are normally distributed with mean 63.6 in and standard deviation 2.5 in. A branch of the military requires​ women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too​ tall? b. If this branch of the military changes the height requirements so that all...
A survey found that women's heights are normally distributed with mean 63.963.9 in and standard deviation...
A survey found that women's heights are normally distributed with mean 63.963.9 in and standard deviation 2.32.3 in. A branch of the military requires women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? b. If this branch of the military changes the height requirements so that all...
Test D: population mean  = 150 and standard deviation  = 20. Danyang’s test score...
Test D: population mean  = 150 and standard deviation  = 20. Danyang’s test score is X = 140. D1.1. Draw the normal curve for this distribution. D1.2. State the value of a score X in the distribution that corresponds to the 99.87th percentile rank. D1.3. Compute the person’s z-score. D1.4. Write an X in the distribution’s horizontal axis where the person’s score is located. D1.5. State the area (proportion) of the curve between the mean  and the...