Thank you so much! <3 -4 points DevoreSt9S.E.058 Ny Rodwal hardness of pns of cc

Question

Thank you so much! <3

-4 points DevoreSt9S.E.058 Ny Rodwal hardness of pns of ccrt in typc is known to havc mean value of 50 2nd stardrd deviation of 1 5. (Round your answers to four dcom! places.) (a) 2f the distribution is normal, what is te probability that the sample mean hardncos forrandam sample of-'1 pins is at least 517 h) what i, theseoranimate) probahility that the sample mean hantriese for a randon, sample of 7 prs is at least 51 cu may reed to use the approoriate table in the Appendix cf Tables to answer this questicn

Explanation / Answer

Mean = 50

Sd = 1.5

Z = (X - mean)/(sd/sqrt(n))

A) n = 14

P(X bar > 51) = P(Z > (51-50)/(1.5/sqrt(14))

= P(Z > 2.49)

= 1 - P(2.49)

= 1 - 0.9936

= 0.0064

B) n = 37

P(Xbar > 51) = P(Z > (51-50)/(1.5/sqrt (37))

= P(Z > 4.06)

= 1 - P(Z < 4.06)

= 1 - almost 1

= 0 (approx)

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