Quiz — corrigé serie proba 2016-2017.pdf

Question 1 sur 10 20:00

[{"id":78796,"question":"Quelle est la formule de l'espérance d'une variable aléatoire X suivant une loi binomiale de paramètres n et p ?","option_a":"E(X) = n + p","option_b":"E(X) = n * p","option_c":"E(X) = p \/ n","option_d":"E(X) = n - p","option_e":"","option_f":"","bonne_reponse":"b","explication":"L'espérance d'une loi binomiale B(n, p) est donnée par E(X) = n * p, où n est le nombre d'essais et p la probabilité de succès.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"E(X) = n + p\", \"b\": \"E(X) = n * p\", \"c\": \"E(X) = p \/ n\", \"d\": \"E(","_debug_options_count":4},{"id":78797,"question":"La variance d'une variable aléatoire suivant une loi normale centrée réduite est égale à 1.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"La variance d'une loi normale centrée réduite est bien égale à 1, car elle est définie par μ = 0 et σ² = 1.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":78798,"question":"Quelle est la probabilité P(X ≤ 1) pour une variable aléatoire X suivant une loi binomiale B(5, 0.2) ?","option_a":"0.32768","option_b":"0.67232","option_c":"0.5","option_d":"0.8","option_e":"","option_f":"","bonne_reponse":"a","explication":"Pour B(5, 0.2), P(X ≤ 1) = P(X=0) + P(X=1) = (0.8)^5 + 5*(0.2)*(0.8)^4 ≈ 0.32768.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"0.32768\", \"b\": \"0.67232\", \"c\": \"0.5\", \"d\": \"0.8\"}}","_debug_options_count":4},{"id":78799,"question":"L'intervalle de confiance pour une proportion p à un niveau de confiance de 95% est donné par :","option_a":"[p - 1.96*σ, p + 1.96*σ]","option_b":"[p - 1.96*√(p(1-p)\/n), p + 1.96*√(p(1-p)\/n)]","option_c":"[p - 2*σ, p + 2*σ]","option_d":"[p - √(p(1-p)\/n), p + √(p(1-p)\/n)]","option_e":"","option_f":"","bonne_reponse":"b","explication":"L'intervalle de confiance pour une proportion p est [p - 1.96*√(p(1-p)\/n), p + 1.96*√(p(1-p)\/n)] pour un niveau de confiance de 95%.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"[p - 1.96*σ, p + 1.96*σ]\", \"b\": \"[p - 1.96*√(p(1-p)\/n), p + 1","_debug_options_count":4},{"id":78800,"question":"La loi normale est une loi de probabilité discrète.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"b","explication":"La loi normale est une loi de probabilité continue, contrairement à la loi binomiale qui est discrète.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":78801,"question":"Quelle est la valeur de l'écart-type σ pour une variable aléatoire X suivant une loi normale N(μ, σ²) ?","option_a":"σ = √(μ)","option_b":"σ = μ","option_c":"σ = √(σ²)","option_d":"σ = μ²","option_e":"","option_f":"","bonne_reponse":"c","explication":"L'écart-type σ est la racine carrée de la variance σ², donc σ = √(σ²).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"c\", \"options\": {\"a\": \"σ = √(μ)\", \"b\": \"σ = μ\", \"c\": \"σ = √(σ²)\", \"d\": \"σ = ","_debug_options_count":4},{"id":78802,"question":"Pour une loi binomiale B(n, p), la variance est donnée par :","option_a":"V(X) = n * p","option_b":"V(X) = n * p * (1 - p)","option_c":"V(X) = p * (1 - p)","option_d":"V(X) = n² * p","option_e":"","option_f":"","bonne_reponse":"b","explication":"La variance d'une loi binomiale B(n, p) est V(X) = n * p * (1 - p).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"V(X) = n * p\", \"b\": \"V(X) = n * p * (1 - p)\", \"c\": \"V(X) = p * (1","_debug_options_count":4},{"id":78803,"question":"Un intervalle de confiance à 99% est plus large qu'un intervalle de confiance à 95% pour la même proportion.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Un intervalle de confiance à 99% est plus large car le coefficient de confiance (2.58) est plus grand que celui à 95% (1.96).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":78804,"question":"Quelle est la probabilité P(0 ≤ X ≤ 2) pour une variable aléatoire X suivant une loi normale N(1, 1) ?","option_a":"0.6826","option_b":"0.9544","option_c":"0.3413","option_d":"0.5","option_e":"","option_f":"","bonne_reponse":"a","explication":"Pour N(1, 1), P(0 ≤ X ≤ 2) ≈ 0.6826 (règle des 68-95-99.7).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"0.6826\", \"b\": \"0.9544\", \"c\": \"0.3413\", \"d\": \"0.5\"}}","_debug_options_count":4},{"id":78805,"question":"La loi binomiale peut être approchée par une loi normale si n est grand et p n'est ni trop proche de 0 ni de 1.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"La loi binomiale peut être approchée par une loi normale si n est grand (n ≥ 30) et si np ≥ 5 et n(1-p) ≥ 5.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4}]