I need help with the last two questions IV & V at the bottom of the page, i have included the other working out to helpa) A company produces metal spacers that are placed between bars in the construction of a range of high precision prefabricated metal items. The spacers are produced so that the length of each is normally distributed with a mean of 40 mm and a standard deviation of 1.5 mm. The spacers are then packaged into groups of 4 for sale. Note: You may use Excel functions to answer the following questions for all parts ii, iii, iv and v and also b. part ii you must provide logical working that includes the Excel function, the Excel output and an appropriate diagram.i) If X is the random variable representing the length of a single spacer, then identify the type of distribution the random variable X has and write down the value(s) of the parameter(s) of this distribution. 1.(i) The distribution of X is Normal distribution.The parameters are Mean, μ = 40Standard deviation, σ = 1.5 ii) Determine the probability that a randomly selected spacer has a length less than 37.75 mm.2.(ii).P( X < 37.75 ) = P[( X - μ )/σ < (37.75 - 40)/1.5] = P( Z < -1.5 ) = 0.06681 Note: The Excel command for getting the p-value of P( Z < -1.5 )=NORM.S.DIST(-1.5,TRUEiii) Calculate the proportion of spacers produced with length greater than 43.75mm.3.(iii):P( X > 43.75 ) = P[( X - μ )/σ > (43.75 - 40)/1.5] = P( Z > 2.5 ) = 0.00621 Note: The Excel command for getting the p-value of P( Z > 2.5 )=1-NORM.S.DIST(2.5,TRUE) iv) Calculate the proportion of spacers produced that have a length within 0.5 mm of the population mean.v) Find the length that 90% of spacers will exceed

Given that. X~N( μ , ?^2 ) μ=40 , ?=1.5 Z-score =( x - μ )/?

Answered: iv) Calculate the proportion of…

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.4: Distributions Of Data

Problem 19PFA

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I need help with the last two questions IV & V at the bottom of the page, i have included the other working out to help

a) A company produces metal spacers that are placed between bars in the construction of a range of high precision prefabricated metal items. The spacers are produced so that the length of each is normally distributed with a mean of 40 mm and a standard deviation of 1.5 mm. The spacers are then packaged into groups of 4 for sale. Note: You may use Excel functions to answer the following questions for all parts ii, iii, iv and v and also b. part ii you must provide logical working that includes the Excel function, the Excel output and an appropriate diagram.

i) If X is the random variable representing the length of a single spacer, then identify the type of distribution the random variable X has and write down the value(s) of the parameter(s) of this distribution.

1.(i) The distribution of X is Normal distribution.

The parameters are

Mean, μ = 40

Standard deviation, σ = 1.5

ii) Determine the probability that a randomly selected spacer has a length less than 37.75 mm.

2.(ii).

P( X < 37.75 )

= P[( X - μ )/σ < (37.75 - 40)/1.5]

= P( Z < -1.5 )

= 0.06681

Note: The Excel command for getting the p-value of P( Z < -1.5 )

=NORM.S.DIST(-1.5,TRUE

iii) Calculate the proportion of spacers produced with length greater than 43.75mm.

3.(iii):

P( X > 43.75 )

= P[( X - μ )/σ > (43.75 - 40)/1.5]

= P( Z > 2.5 )

= 0.00621

Note: The Excel command for getting the p-value of P( Z > 2.5 )

=1-NORM.S.DIST(2.5,TRUE)

iv) Calculate the proportion of spacers produced that have a length within 0.5 mm of the population mean.
v) Find the length that 90% of spacers will exceed

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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