Question

Find the probability that a piece of data from a standard normal distribution will have a standard score described by the fol
A universitys administrator proposes to do an analysis of the proportion of graduates who have not found employment in their
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Answer #1

1).a).\mathbf{P(Z<1.25)}

= Φ(1.25)

\approx\mathbf{0.8944} [ using normal table]

b).\mathbf{P(-2.03<Z<-0.69)}

= P(Z < -0.69) - P(Z < -2.03)

=\Phi(-0.69)-\Phi(-2.03)

=1 – Φ(0.69) – 1+ Φ(2.03)

= -Φ(0.69) + Φ(2.03)

=-0.7549+0.9788 [ using normal table]

=\mathbf{0.2239}

c).let z be the 62nd percentile.

P(Z\leq z)=0.62

\Rightarrow \Phi( z)=0.62

\Rightarrow z=\Phi^{-1}(0.62)

\Rightarrow z=0.3055\approx\mathbf{0.31}

the score for 62nd percentile = 0.31

2).the given data are:-

margin of error (E) = 2% = 0.02

proportion (p) = 0.12

z critical value for 92% confidence level, both tailed test be:-

z^*=1.751 [ in excel type =NORMSINV(0.96)]

the needed sample size be:-

=\left ( \frac{z^*}{E} \right )^2*p*(1-p)

=\left ( \frac{1.751}{0.02} \right )^2*0.12*(1-0.12)

\approx\mathbf{810}

*** if you have any doubt regarding the problem please write it in the comment box.if you are satisfied please give me a LIKE if possible...

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