A manufacturer knows that their items have a normally distributed length, with a mean of 17.7 inches, and standard deviation of 5.2 inches. If 22 items are chosen at random, what is the probability that their mean length is less than 14.7 inches?

Q: A Texas Instrument calculator operating life is normally distributed with a mean of 3,850 hours, and…

A: Given information- Population mean, μ = 3850 hours Population standard deviation, σ = 350 hours…

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 14.2…

A: GivenMean(μ)=14.2standard deviation(σ)=2.2

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 14.1…

A:

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 15.4…

A: We have to find the probability that their mean length is less than 15.9 inches...

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 14.6…

A:

Q: A particular fruit's weights are normally distributed, with a mean of 557 grams and a standard…

A: From the given information, let X be the fruit’s weights. Here, X follows normally distribution with…

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 18.6…

A: Suppose the random variable x defines the length of the items.

Q: A final exam in Math150 has a mean of 73 with standard deviation 7.8. If 24 students are randomly…

A:

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 5.1 inches,…

A: The provided information is as follows:The population mean is  inches.The standard deviation is…

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 14.5…

A:

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 13.2…

A: The random variable is item. It follows standard normal distribution. It is given that the…

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 7.1 inches,…

A: Mean(µ) = 7.1Standard deviations (σ) = 2.2X ~ N (µ, σ )= N (7.1, 2.2)

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 9.1 inches,…

A: It is given that the mean is 9.1 and standard deviation is 1.3.

Q: manufacturer knows that their items have a normally distributed length, with a mean of 15.6 inches,…

A: Provided information is, a manufacturer knows that their items have a normally distributed length…

Q: The average number of pounds of red meat a person consumes each year is 196 with a standard…

A: We are given,μ = 196σ = 22n = 60

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 5.4 inches,…

A: From the provided information, Mean (µ) = 5.4 Standard deviation (σ) = 1.5

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 7.4 inches,…

A: Let "M" be the mean length of the item in inches.

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 17.4…

A:

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 12.4…

A:

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 12.4…

A: Given Information: Mean μ=12.4 Standard deviation σ=2.6 If 4 items are chosen at random, to find the…

Q: manufacturer knows that their items have a lengths that are skewed right, with a mean of 19.3…

A: Given that.   X~N( μ , ?^2 )   μ=19.3     , ?=5.7  ,n=48   Z-score =( x - μ )/?

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 10.8…

A: Given that  Mean = M = 10.8  Standard deviation = SD = 1.6 Sample size = n = 4 X ~ Normal (M = 10.8…

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 19.3 inches…

A: It is given that the mean is 19.3 and the standard deviation is 2.

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 15.4…

A: We know it is normally distributed. So we can use the z - probability formula. z=x-μσ Here, μ -…

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 12.1…

A:

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 14.7…

A:

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 15 inches,…

A: Mean = 15 Standard deviation = 1

Q: Blood pressure is distribued Normally, with a mean of 85 mm and a standard deviation of 20 mm. As a…

A:

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 16.5…

A: Central Limit Theorem for mean: If a random sample of size n is taken from a population having mean…

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 9.2 inches,…

A: The distribution of length of items is N(µ=9.2 inches, σ=1.6 inches).

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 12.4…

A: Mean = 12.4 Standard deviation = 1.7

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 5.5 inches,…

A:

Q: A manufacturer knows that their items have a normally distributed length, with a mean of 14.5…

A: We have to find given probability.

Q: The height of 10 year old females is normally distributed with a mean of 54.8 inches and a standard…

A: It is given that mean and standard deviation are 54.8 inches and 2.74 inches, respectively.

Q: at their items have a normally distributed length, with a mean of 16.8 inches, and standard…

A: According to the sum, A manufacturer knows that its items have a normally distributed length, with a…

Q: Suppose that heights of women at UT Tyler are normally distributed with a mean of 64 inches and a…

A: It is given that the mean is 64 and standard deviation is 2.4.

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.