Solusyon - Pangkabuuang probabilidad sa standard normal na distribusyon

Pangkabuuang probabilidad

84.134 %

84.134%

Tingnan ang mga hakbang

Hakbang-sa-hakbang na paliwanag

1. Alamin ang pangkabuuang probabilidad ng mga halaga ng z-score hanggang sa $1$

Gamitin ang positive z-table upang mahanap ang halaga na tumutugma sa $1$ . Ang halagang ito ay ang pangkabuuang probabilidad ng area sa kaliwa ng $1$ .

Z	0.00	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
0.0	0.5	0.50399	0.50798	0.51197	0.51595	0.51994	0.52392	0.5279	0.53188	0.53586
0.1	0.53983	0.5438	0.54776	0.55172	0.55567	0.55962	0.56356	0.56749	0.57142	0.57535
0.2	0.57926	0.58317	0.58706	0.59095	0.59483	0.59871	0.60257	0.60642	0.61026	0.61409
0.3	0.61791	0.62172	0.62552	0.6293	0.63307	0.63683	0.64058	0.64431	0.64803	0.65173
0.4	0.65542	0.6591	0.66276	0.6664	0.67003	0.67364	0.67724	0.68082	0.68439	0.68793
0.5	0.69146	0.69497	0.69847	0.70194	0.7054	0.70884	0.71226	0.71566	0.71904	0.7224
0.6	0.72575	0.72907	0.73237	0.73565	0.73891	0.74215	0.74537	0.74857	0.75175	0.7549
0.7	0.75804	0.76115	0.76424	0.7673	0.77035	0.77337	0.77637	0.77935	0.7823	0.78524
0.8	0.78814	0.79103	0.79389	0.79673	0.79955	0.80234	0.80511	0.80785	0.81057	0.81327
0.9	0.81594	0.81859	0.82121	0.82381	0.82639	0.82894	0.83147	0.83398	0.83646	0.83891
1.0	0.84134	0.84375	0.84614	0.84849	0.85083	0.85314	0.85543	0.85769	0.85993	0.86214
1.1	0.86433	0.8665	0.86864	0.87076	0.87286	0.87493	0.87698	0.879	0.881	0.88298
1.2	0.88493	0.88686	0.88877	0.89065	0.89251	0.89435	0.89617	0.89796	0.89973	0.90147
1.3	0.9032	0.9049	0.90658	0.90824	0.90988	0.91149	0.91308	0.91466	0.91621	0.91774
1.4	0.91924	0.92073	0.9222	0.92364	0.92507	0.92647	0.92785	0.92922	0.93056	0.93189
1.5	0.93319	0.93448	0.93574	0.93699	0.93822	0.93943	0.94062	0.94179	0.94295	0.94408
1.6	0.9452	0.9463	0.94738	0.94845	0.9495	0.95053	0.95154	0.95254	0.95352	0.95449
1.7	0.95543	0.95637	0.95728	0.95818	0.95907	0.95994	0.9608	0.96164	0.96246	0.96327
1.8	0.96407	0.96485	0.96562	0.96638	0.96712	0.96784	0.96856	0.96926	0.96995	0.97062
1.9	0.97128	0.97193	0.97257	0.9732	0.97381	0.97441	0.975	0.97558	0.97615	0.9767
2.0	0.97725	0.97778	0.97831	0.97882	0.97932	0.97982	0.9803	0.98077	0.98124	0.98169
2.1	0.98214	0.98257	0.983	0.98341	0.98382	0.98422	0.98461	0.985	0.98537	0.98574
2.2	0.9861	0.98645	0.98679	0.98713	0.98745	0.98778	0.98809	0.9884	0.9887	0.98899
2.3	0.98928	0.98956	0.98983	0.9901	0.99036	0.99061	0.99086	0.99111	0.99134	0.99158
2.4	0.9918	0.99202	0.99224	0.99245	0.99266	0.99286	0.99305	0.99324	0.99343	0.99361
2.5	0.99379	0.99396	0.99413	0.9943	0.99446	0.99461	0.99477	0.99492	0.99506	0.9952
2.6	0.99534	0.99547	0.9956	0.99573	0.99585	0.99598	0.99609	0.99621	0.99632	0.99643
2.7	0.99653	0.99664	0.99674	0.99683	0.99693	0.99702	0.99711	0.9972	0.99728	0.99736
2.8	0.99744	0.99752	0.9976	0.99767	0.99774	0.99781	0.99788	0.99795	0.99801	0.99807
2.9	0.99813	0.99819	0.99825	0.99831	0.99836	0.99841	0.99846	0.99851	0.99856	0.99861
3.0	0.99865	0.99869	0.99874	0.99878	0.99882	0.99886	0.99889	0.99893	0.99896	0.999
3.1	0.99903	0.99906	0.9991	0.99913	0.99916	0.99918	0.99921	0.99924	0.99926	0.99929
3.2	0.99931	0.99934	0.99936	0.99938	0.9994	0.99942	0.99944	0.99946	0.99948	0.9995
3.3	0.99952	0.99953	0.99955	0.99957	0.99958	0.9996	0.99961	0.99962	0.99964	0.99965
3.4	0.99966	0.99968	0.99969	0.9997	0.99971	0.99972	0.99973	0.99974	0.99975	0.99976
3.5	0.99977	0.99978	0.99978	0.99979	0.9998	0.99981	0.99981	0.99982	0.99983	0.99983
3.6	0.99984	0.99985	0.99985	0.99986	0.99986	0.99987	0.99987	0.99988	0.99988	0.99989
3.7	0.99989	0.9999	0.9999	0.9999	0.99991	0.99991	0.99992	0.99992	0.99992	0.99992
3.8	0.99993	0.99993	0.99993	0.99994	0.99994	0.99994	0.99994	0.99995	0.99995	0.99995
3.9	0.99995	0.99995	0.99996	0.99996	0.99996	0.99996	0.99996	0.99996	0.99997	0.99997

Ang z-score ng $1$ ay tumutugma sa isang lugar ng $0.84134$
$p (z < 1) = 0.84134$
Ang nakuhang peluang na $z < 1$ ay may $84.134 %$

2. Alamin ang pangkabuuang probabilidad ng mga halaga ng z-score hanggang sa $- 13.333$

Mahigit sa $99.9 %$ ng oras, ang datos na may pamantayang normal na pamamahagi ay nasa loob ng dagdag o bawas na $3.9$ na pamantayang dibeyansya ng mean.

Ang pampamahaging katotohanan ng mga halaga hanggang sa $- 13.333$ ay $0$ .
$p (z < - 13.333) = 0$
Ang pampamahaging katotohanan na $z < - 13.333$ ay $0 %$

3. Kalkulahin ang pangkabuuang probabilidad sa pagitan ng 1 at -13.333

Para mahanap ang cumulative probability ng lugar sa pagitan ng dalawang z-scores, ibawas ang mas maliit na cumulative probability (lahat sa kaliwa ng $- 13.333$ ) mula sa mas malaking cumulative probability (lahat sa kaliwa ng $1$ ):

$0.84134 - 0 = 0.84134$
$p (- 13.333 < z < 1) = 0.84134$
Ang cumulative probability na $- 13.333 < z < 1$ ay $84.134 %$

Paano tayo nagtrabaho?

Mangyaring mag-iwan ng iyong puna

Bakit kailangan matutuhan ito

Bakit matutunan ito

Mga Terminolohiya at Paksa

Karaniwan at pamantayang normal na pamamahagi

Solusyon - Pangkabuuang probabilidad sa standard normal na distribusyon

Hakbang-sa-hakbang na paliwanag

1. Alamin ang pangkabuuang probabilidad ng mga halaga ng z-score hanggang sa 1

2. Alamin ang pangkabuuang probabilidad ng mga halaga ng z-score hanggang sa −13.333

3. Kalkulahin ang pangkabuuang probabilidad sa pagitan ng 1 at -13.333

Bakit kailangan matutuhan ito

Mga Terminolohiya at Paksa

Mga kaugnay na link

Pinakabagong Kaugnay na mga Drills na Nalutas

1. Alamin ang pangkabuuang probabilidad ng mga halaga ng z-score hanggang sa $1$

2. Alamin ang pangkabuuang probabilidad ng mga halaga ng z-score hanggang sa $- 13.333$