For standard normal random variable Z, determine the value of constant c which makes the probability statements given below correcta) Φ ( c ) = 0.8888b) P ( 0 ≤ Z ≤ c ) = 0.334c) P ( − c ≤ Z ≤ c ) = 0.488e) P ( c ≤ | Z | ) = 0.154

Question

For standard normal random variable Z, determine the value of constant c which makes the probability statements given below correcta)   Φ  (  c  )  =  0.8888b)   P  (  0  ≤  Z  ≤  c  )  =  0.334c)   P  (  −  c  ≤  Z  ≤  c  )  =  0.488e)   P  (  c  ≤      |    Z      |    )  =  0.154

Zackary Galloway · Accepted Answer

Step 1a)   Φ  (  c  )  =  0.8888    ⇒  P  (  Z  ≤  c  )  =  0.8888From standard norma tables, the value corresponding to 0.8888 cumulative probability is 1.22Hence,   c  =  1.22Step 2b)   P  (  0  ≤  Z  ≤  c  )  =  0.334    ⇒  P  (  Z  ≤  c  )  −  P  (  Z  ≤  0  )  =  0.334    ⇒  P  (  Z  ≤  c  )  −  0.5  =  0.334    (  since     P  (  Z  ≤  0  )  =  P  (  Z  ≥  0  )  =  0.5  )    ⇒  P  (  Z  ≤  c  )  =  0.834From standard normal tables, the value corresponding to 0.8888 cumulative probability is 0.97Hence,   c  =  0.97

Kody Whitaker · Accepted Answer

MathJax(?): Can&#039;t find handler for document MathJax(?): Can&#039;t find handler for document Step 1c)   P  (  c  &amp;#x2264;  Z  )  =  0.271    &amp;#x21D2;  P  (  Z  &amp;#x2265;  c  )  =  0.271    &amp;#x21D2;  1  &amp;#x2212;  P  (  Z  &amp;lt;  c  )  =  0.271    &amp;#x21D2;  P  (  Z  &amp;lt;  c  )  =  1  &amp;#x2212;  0.271    &amp;#x21D2;  P  (  Z  &amp;lt;  c  )  =  0.729From standard norma tables, the value corresponding to 0.729 cumulative probability is 0.61Hence,   c  =  0.61d)   P  (  &amp;#x2212;  c  &amp;#x2264;  Z  &amp;#x2264;  c  )  =  0.488    &amp;#x21D2;  P  (  Z  &amp;#x2264;  c  )  &amp;#x2212;  P  (  Z  &amp;#x2264;  &amp;#x2212;  c  )  =  0.488    &amp;#x21D2;  P  (  Z  &amp;#x2264;  c  )  &amp;#x2212;  P  (  Z  &amp;gt;  c  )  =  0.488  &amp;#xA0;  (Since Z follows normal distribution which is symmetric)    &amp;#x21D2;  P  (  Z  &amp;#x2264;  c  )  &amp;#x2212;  [  1  &amp;#x2212;  P  (  Z  &amp;#x2264;  c  )  ]  =  0.488    &amp;#x21D2;  2  P  (  Z  &amp;#x2264;  c  )  &amp;#x2212;  1  =  0.488    &amp;#x21D2;  2  P  (  Z  &amp;#x2264;  c  )  =  1.488    &amp;#x21D2;  P  (  Z  &amp;#x2264;  c  )  =  0.744&amp;#x21D2;

For standard normal random variable Z, determine the value of constant c which makes the probability statements given below correct

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