## Trending News

# 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

- AlanLv 72 years agoFavorite Answer
(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

- Login to reply the answers