Probability Question! Statistics Please help!?

The breaking strength (in kg/mm) for a certain type of fabric has mean 1.86 and standard

deviation 0.27. A random sample of 80 pieces of fabric is drawn.

(a) What is the probability that the sample mean breaking strength is less than 1.8 kg/mm? Show all the

steps. What theorem do you apply to answer this question?

(b) Find the 80th percentile of the sample mean breaking strength. Show all the steps

(c) How large a sample size is needed so that the probability is 0.01 that the sample mean is less than 1.8?

Show all the steps

1 Answer

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  • Alan
    Lv 7
    2 years ago
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    (a) What is the probability that the sample mean breaking strength is less than 1.8 kg/mm? Show all the

    steps. What theorem do you apply to answer this question?

    Based on the Central Limit Thereom

    for the Sample mean

    mean = population mean

    Standard Deviation = population standard deviaton / sqrt(N) where is the number of samples

    mean = 1.86

    Std dev for sample mean = 0.27/ sqrt(80) = 0.030186918

    since Z = (x- mean) / standard deviation of sample mean

    P(X_bar <1.8) = P( Z < (1.8-1.86) / 0.030186918 ) = P(Z < -1.98761598 )

    P(X_bar< 1.8) = P(Z < -1.98761598 )

    From this table at link below

    https://www.stat.tamu.edu/~lzhou/stat302/standardn...

    P(Z< -1.99) = .02385

    P(Z< -1.98) = .02330

    Interpolating for -1.98761598

    P = 0.02385 + (-1.98761598- (-1.99) ) *( 0.02330 -0.02385) / (0.01) =0.023718879

    rounding to 5 digits since the table is only accurate to 5 digits

    P(X_bar < 1.8 kg/mm) = 0.02372

    (b) Find the 80th percentile of the sample mean breaking strength. Show all the steps

    First from the P(z< Z) = 0.80

    two closest vaue from the table

    P(Z< 0.84) = .79955

    P(Z< 0.85) = .80234

    Interpolating

    0.84 + ( 0.80 - 0.79955 ) * (0.01) / ( 0.80234 - 0.79955)

    = 0.841612903

    (rounding to 3 digits

    = 0.842

    now using this with Z formula

    Z_sample_mean = ( x - mean) / (standard deviation_Sample_mean)

    (x- mean) =Z_sample_mean*standard_Deviaton_sample_mean

    x = mean + Z_sample_mean*standard_Deviation_Sample_mean

    x = 1.86 + 0.842 * 0.030186918

    ===== Answer to (b) round to the appropriate digits

    x = 1.885417385

    (c)

    so you P(z< Z) = 0.01

    so look two closest values

    P(z< -2.33 ) = . .00990

    P(z< -2.32) = 01017

    interpolate

    -2.33 + (0.01 - 0.0099) * 0.01 / (0.01017 - 0.0099) =

    =-2.326296296

    rounding to 3 digits

    = -2.326

    We can use this same formula we found before

    but

    Z = (x-standard deviation) / (standard_dev/ sqrt(N)

    -2.326 = (1.8-1.86) / (0.27/sqrt(N) )

    0.27*-2.326/sqrt(N) = -0.06

    -0.06*sqrt(N) = 0.27*-2.326

    sqrt(N) = 0.27*(-2.326) / -0.06

    sqrt(N) = 10.467

    square both sides

    N = 109.558089

    but N can have a fraction ,

    N = 109.558089 is number of sample which would make exactly 0.01, but of couse

    it is hard to take 0.558089 of a sample.

    This question is terriblely worded.

    To pick an N , it needs to such that probability is less than 0.01 or greater than 0.01

    It will never be exactly 0.01

    If N = 109

    Z = -0.06/ (0.27/sqrt(109) = -2.320068113 (greater than -2.326)

    so P(x<1.8) = approx. 0.010168596

    so for N = 109 , the probability is greater than 0.01 (just barely)

    If N =110

    Z = -0.06/ (0.27/sqrt(109) = -2.330686329 (less -.2.326)

    P(z< -2.3306) = 0.009884953

    so 109 makes it greater than 0.01

    so 110 make it less than 0.01

    No value of N (because it can only be an integer) will make it exactly equal

    So if you reworded the question to

    how big of a sample size is needed such that the probability is less than 0.01 for the sample mean

    to be less than 1.8 , the answer is 110

    === expected answer I think you want

    N = 110

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