Stat 251 Exam 2 makeup Section 02/05 20 October 9017 1. The Midgur Department of

Question

Stat 251 Exam 2 makeup Section 02/05 20 October 9017 1. The Midgur Department of Commerce, run by Shiers Corporation, sreports that the mn temperature in dowstows Midgat (ahove the Plate) is 65.3 F with standued deviaticn 78F and follows a normal distribatione (Le XN(3,78). What is the probshlity of a raoely solectod day with tempera ture more than 72 F (a) 0.8048 (b) 0.1587 (c) 0.1932 (d) o (e) 1 2. Uning theidination feom #1, what t mperatu”-the-mont 25%of dys? (a) 512F () 706 F (c) 0.03 (d) 002° F hours. Let X the name of this distribution? Whst iyare the persetor (b) Normal -9and -10 (d) Goethetric,.. 9 )? (e) none of the above in the next ten seconds (a) 1 (b) 00025 le) 0 9999 (d) 0.0001 (e)0.8413 5. Using the information froms e3, what is the probillay that at lesst ose car antives in the sext ten soconds? (a) 0 (b) 0.0001 (c) 0.9975 (d) 0.1587 (e) 0.9999

Explanation / Answer

1) Z= (72-65.3)/7.8 = 0.85

P(Z>0.85) = 0.1952 from normal distribution tables.

2) Z for cumulative probability = 0.75 is 0.67449

Temperature = 65.3+0.67449*7.8= 70.56.

Hence, B is the answer

Please repost the other questions individually.

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